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GIFT  OF 

U,C.  Lambda  Chapter 

Phi  Delta  Kappa 


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ARITHMETIC 


AND   ITS  APPHC^f  ib'ks  *;      '     ' 


*  »  ;  -• 


>  >     1    • 


DESIGNED   AS    A 


TEXT  BOOK  POK  COMMON  SCHOOLS, 


HIGH   SCHOOLS,   AND   ACADEMIES'. 


BY 

DANA    P.    COLBURN, 

PSINCIPAL    OP    THE     RHODE     ISLAND    STATE     NORMAL    SCITOOI,, 
PROVIDENCB. 


PHILADELPHIA: 

H.    COWPERTHWAIT    &    CO. 

BOSTON:  SHEPARD,  CLARK  &  CO. 

1858. 


Entered,  according  to  Act  of  Ctongrwg,  In  the  Year  1866.  by 

DANA    P     COLBCRN, 

In  the  Clerk's  Office  of  the  District  Court  of  the  District  of  Rhode  iRlanO. 


PRTVTKD  HV  !5VTTTf  *  PTTTRK^ 


PRE  PACE. 


The  principles  involved  in  Arithmetic  are  few,  the  meth- 
ods of  applying  them  many.  To  be  a  perfect  master  of  the 
subject,  a  person  must  possess,  — 

1.  A  knowledge  of  the  nature  and  use  of  numbers,  with 
the  methods  of  representing  and  expressing  them. 

2.  A  knowledge  of  the  nature  and  use  of  the  various 
numerical  operations,  with  the  methods  of  indicating  and  of 
performing  them. 

3.  Such  mental  training  and  cultivation  of  the  reasoning 
powers  as  shall  enable  him  to  understand  the  conditions  of 
any  given  problem,  and  to  determine  from  them  what  opera- 
tions are  necessary  to  its  solution. 

,  The  first  of  these  includes  every  thing  belonging  to  Nota- 
tion and  Numeration. 

The  second  includes  the  operations  of  Addition,  Subtraction, 
Multiplication,  and  Division ;  to  which  some  would  add,  as  a 
separate  operation,  the  Comparison  of  Numbers,  as  in  Frac- 
tions and  Ratio. 

The  third  requires  a  power  of  grasping  the  various  condi- 
tions of  a  problem,  of  tracing  their  relations  to  each  other,  and 
of  finding  from  them  what  operations  must  be  performed,  and 
what  new  relations  determined,  to  obtain  the  result  required. 

They  are,  however,  mutually  dependent,  so  that  no  person 


hr  PREFACE. 

can  master  one  without  learning  much  of  the  others.  Count* 
ing  is  but  addition ;  and  to  understand  the  nature  and  use  of 
the  number  two,  we  must  know  that  it  equals  1  and  1,  two 
ones,  or  two  times  1 ;  that  it  is  1  more  than  1 ;  that  if  1  be 
taken  from  it  1  will  be  left,  and  so  on  with  the  other  num- 
bers —  things  which  require  a  knowledge  of  numerical  opera- 
tions, and  also  a  power  of  tracing  and  appreciating  relations. 

The  operations  and  exercises  included  in  the  first  two  points 
are  eminently  adapted  to  give  quickness  of  thought  and  rapid- 
ity of  mental  action.  They  are  to  Arithmetic  what  a  knowl- 
edge of  the  nature  and  power  of  letters,  and  of  their  combina- 
tion into  words,  is  to  reading.  The  processes  included  in  them 
may  be  called  the  mechanical  processes  of  Arithmetic,  and 
by  practice  may  and  should  be  made  so  familiar  that  the  mo- 
ment a  number  or  a  combination  is  suggested,  the  mind  can 
appreciate  it,  and  determine  the  result. 

The  third  requires  and  imparts  a  power  of  investigation, 
of  tracing  out  the  relations  of  cause  and  effect,  and  habits  of 
accuracy  both  in  thought  and  expression. 

To  secure  these  results,  it  is  necessary  that  the  pupil  should 
be  taught  in  the  simplest  as  well  as  in  the  most  complicated 
problems  to  reason  for  himself;  to  trace  fully  and  clearly  the 
connection  between  the  conditions  of  a  problem  and  the  steps 
taken  in  its  solution  ;  to  stale  not  only  what  he  does,  but  why 
he  does  it,  and  indicate  the  precise  character  of  the  result 
obtained  by  each  step.  Finally,  he  must  learn  to  grasp  the 
whole  mechanical  process  before  performing  any  part  of  it,  so 
that  he  may  know  before  writing  a  figure  just  what  additions, 
subtractions,  multiplications,  divisions,  and  comparisons  he  has 
to  make,  and  be  assured  that  if  made  correctly  they  will  lead 
to  the  true  result. 

Such  a  course  as  this  is  usually  taken  in  works  on  Oral 
Arithmetic.     In  studying  them  tlie  scholar  is  thrown  on  his 


PREFACE. 


own  resources;  is  compelled  to  learn  principles  ;  to  folic w  out 
rigid  reasoning  processes  and  connected  trains  of  thought ;  to 
examine  and  know  for  himself  the  necessity  and  the  reason  for 
each  step  taken,  and  for  each  operation  performed.  The  result 
is,  that  the  study  gives  strength,  vigot,  and  healthful  disci- 
pline to  the  mind,  and  becomes  an  almost  invaluable  part  of 
the  educating  process. 

Why  should  not  the  same  result  follow  a  similar  course  in 
Written  Arithmetic?  Aside  from  the  writing  of  numbers, 
there  is  no  difference  in  the  principles  involved,  in  the  reason- 
ing processes  demanded,  or  in  the  operations  required. 

In  the  preparation  of  this  work,  the  author  has  kept  these 
things  in  view.  He  has  endeavored  to  present  the  subject  of 
Arithmetic  as  it  lies  in  his  own  mind,  and  without  any  effort 
either  to  follow  or  t  ^  deviate  from  the  course  pursued  by  other 
writers.  He  has  aimed  to  arrange  the  work  in  such  a  way  as 
to  lead  those  who  may  study  it  to  understand  the  principles 
which  lie  at  the  foundation  of  the  science,  to  learn  to  reason 
upon  them,  apply  them,  and  to  trace  out  their  connections, 
relations,  and  combinations.  He  has  given  very  full  explana- 
tions and  illustrations,  especially  of  the  fundamental  opera- 
tions ;  he  has  endeavored  every  where  to  state  principles 
rather  than  rules ;  to  throw  the  pupil  constantly  on  his  own 
resources,  and  force  him  to  investigate  and  think  for  himself 

He  has  omitted  some  subjects  usually  found  in  school  arith- 
metics, because  they  do  not  belong  legitimately  to  the  subject 
of  Arithmetic,  because  they  are  of  theoretical  rather  than  of 
practical  importance,  or  because  they  require  neither  special 
explanation  nor  peculiar  exercise  of  the  mind. 

He  has  differed  from  other  authors  of  school  arithmetics  in 

giving  algebraic  rather  than  geometrical  explanations  of  the 

principles  involved  in  Square  and  Cube  Roots.     In  this  way 

he  has  been  able  to  give  moro  rigid  demonstrations,  and  more 

o* 


fi  PREFACE. 

full  explau&tions,  and  at  the  same  time  (as  he  conceives)  to 
simplify  the  subject.  Moreover  the  processes  are  in  their 
nature  so  essentially  algebraic  that  by  the  use  of  squares  and 
cubical  blocks  we  can  do  nothing  more  than  illustrate  some 
of  their  applications. 

He  has  given  no  answers  to  his  problems,  because  he  be- 
lieves that  to  place  them  within  reach  of  the  pupil  is  always 
injurious. 

In  the  first  place,  such  tests  are  unpractical,  for  they  can 
never  be  resorted  to  in  the  problems  of  real  life.  What  mer- 
chant ever  thinks  of  looking  in  a  text  book  or  a  key,  or  of  rely- 
ing on  his  neighbor,  to  learn  whether  he  has  added  a  column 
correctly,  drawn  a  correct  balance  between  the  debit  and 
credit  sides  of  an  account,  or  made  a  mistake  in  finding  the 
amount  of  a  bill  ? 

When  a  pupil,  having  left  the  school  room,  performs  a  prob- 
lem of  real  life,  how  anxious  is  he  to  know  whether  his  result 
be  correct !  Neither  text  book  nor  key  can  aid  him  now,  and 
he  is  forced  to  rely  on  himself  and  his  own  investigations  to 
determine  the  truth  or  the  falsity  of.his  work.  If  he  must  al- 
ways do  this  in  real  life,  and  if  his  school  course  is  to  be  a 
preparation  for  the  duties  of  real  life,  ought  he  not  to  do  it  as 
a  learner  in  school  ?  Is  it  right  to  lead  him  to  rely  on  such 
false  tests  ? 

Besides,  the  labor  of  proving  an  operation  is  usually  as  val- 
uable arithmetical  work  as  was  the  labor  of  performing  it,  and 
will  oftentimes  make  a  process  or  solution  appear  perfectly 
simple  and  clear,  when  it  would  otherwise  have  seemed  ob- 
scure and  complicated. 

Again :  the  science  of  Mathematics,  of  which  Arithmetic  is 
a  branch,  is  an  exact  science  ;  it  deals  in  no  uncertainties  ;  its 
reasonings  are  always  accurate,  and,  if  based  on  true  prem- 
ises, must  always  lead  to  trije  resplts.     In  Arithmetic  the 


PREFACE.  VU 

pupil  may  always  know  that  a  certain  step  is  a  true  one,  and 
one  which  he  has  a  right  to  take.  He  may  know  whether  he 
has  taken  it  correctly,  and  thus  be  certain  of  the  truth  of  his 
first  result.  He  may  be  as  sure  of  the  truth  of  his  second 
step  and  second  result,  and  of  his  third  and  his  fourth;  and 
when  he  reaches  the  end,  and  obtains  his  final  result,  he  may 
be  as  sure  of  the  truth  of  that  as  of  any  preceding  —  so  sure 
that  he  will  be  willing  to  abide  by  it,  and  stake  his  reputation 
upon  it.     (See  page  49.) 

Why,  then,  should  not  the  subject  be  so  presented  as  to 
require  the  pupil  to  apply  such  tests,  to  determine  for  himself 
the  truth  and  accuracy  of  his  processes,  and  thus  to  form  a 
habit  of  patient  investigation  and  just  self-reliance  ?  Why 
should  he  not  be  from  the  first  thrown  on  his  own  resources, 
and  held  strictly  responsible  for  the  accuracy  of  his  work  ? 
Would  not  such  a  course,  if  faithfully  followed,  almost  entirely 
prevent  the  formation  of  those  careless  habits  which  scholars 
so  often  acquire  ? 

The  articles  on  business  forms  and  transactions  have  been 
carefully  prepared,  with  a  hope  of  so  presenting  them  as  to 
give  the  student  true  ideas  of  their  use,  and  of  the  relations 
and  obligations  of  the  parties  to  them. 

The  materials  were  drawn  from  various  sources  —  from 
legal  works,  from  intercourse  with  business  men,  and  from  an 
article  in  Mann  and  Chase's  Arithmetic,  published  originally 
in  the  Common  School  Journal.  To  insure  accuracy  they 
were  submitted  to  the  inspection  of  Abraham  Payne,  Esq.,  an 
eminent  lawyer  of  this  city,  to  whom  I  am  indebted  for  some 
important  suggestions. 

The  work  as  a  whole  resembles  all  other  text  books  (good 
or  bad)  in  this  —  that  it  requires  a  good  teacher  to  teach  it 
well ;  as  also  in  this  —  that  it  does  not  contain  exactly  the 
right  kind  and  amount  of  exercises  to  meet  the  wants  of  every 


▼Ill  PREFACE. 

school  or  of  every  class  of  scholars.  The  judicious  teacher 
will  of  course  extend  the  exercises  which  are  too  meagre, 
abridge  those  which  are  too  full,  and  omit  those  which  are  not 
adapted  to  the  wants  of  his  class.  We  earnestly  beg  of  him, 
however,  to  notice  their  arrangement,  their  gradual  chai-- 
acter,  and  their  dependence  on  each  other ;  and  not  to  pass 
any  till  he  has  convinced  himself  that  they  are  inappropri- 
ate, or  that  the  scholar  is  master  of  the  operations  which  they 
involve. 

To  my  former  teacher,  N.  Tillinghast,  Esq.,  for  many  years 
principal  of  the  State  Normal  School  at  Bridgewater,  Massa- 
chusetts, I  am  more  deeply  indebted  than  to  any  other,  or  all 
others,  for  the  ideas  embodied  in  this  work.  Many  of  the 
processes  were  learned  under  his  tuition ;  and  the  training 
which  laid  the  foundation  for  whatever  real  mathematical 
knowledge  I  may  possess,  was,  in  a  great  measure,  received 
from  him.  Only  those  who  have  been  his  pupils  can  appre- 
ciate the  value  of  his  instructions,  and  the  justice  of  this 
acknowledgment. 

The  work  in  its  plan  and  arrangement  is  entirely  my  own, 
and  for  its  defects  I  alone  must  be  held  responsible.  Such  as 
it  is  I  present  it  to  the  public,  with  a  hope  that  it  may  be 
found  useful. 

DANA  P.  COLBURN. 

F&OVIDBNCB,  Julut  1855. 


CONTENTS. 


SECTION  L 


Deflnitlons, 


PAGE 
.      1 


CI.  NOTA'JION  AND  NUMERATION 


Definitions  of  Tenns,    .... 

Origin  of  the  Decimal  System,     . 

Nature  of  the  Decimal  System,    . 

Arabic  and  Roman  Methods, 

Figures, 

Denomination  determined  by  Place,    . 

Names  and  P;,sition  of  Decimal  Places, 

Method  of  reading  Numbers, 

Exercises  in  writing  and  reading  Num- 
bers,     

Number  of  Decimal  Places  unlimited. 

Division  into  Periods,    .... 

Exercises  on  Places  and  Periods,  . 

Method  of  reading  Numbers, 

All  intervening  Places  to  be  filled, 

Method  of  writing  Numbers, 

Any  Combination  may  be  read  by  itself. 

Analysis,  of  Numbers,   .        .        .        . 

Comparison  of  Figures  in  different 
Places,        

Places  at  Right  of  Point, 

Method  of  reading  Decimal  Fractions, 

Method  of  writing  Decimal  Fractions, 

Exercises  in  determining  Position  of 
Point, 

Effect  of  changing  Place  of  Point, 

Change  of  Denomination,    . 

French  Method  of  Numeration,  . 

English  Method  of  Numeration,  . 

Comparison  of  Methods, 

Exercises,      ... 

Reman  Method. 


III.  TABLES. 

Introductory, 

27 

United  States  Money,   . 

27 

Exercises,       .... 

.  28 

English  Money,     . 

.  28 

Avoirdupois  Weight,    . 

.        .  29 

Troy  Weiglit, 

.        .30 

Apothecaries'  Weight, . 

.        .  30 

Comparison  of  Weights, 

.  31 

Long  Measure, 

.  32 

Cloth  Measure,      .       , 

.        .  32 

Square  Measure,  .        , 

.        .33 

Cubic  Measure,    . 

.  35 

Angular  Measure, . 

.  36 

Dry  Measure, 

38 

Liquid  Measure,    . 

.38 

Comparison  of  Measures,     . 

.39 

Table  of  Time,      . 

.  39 

Miscellaneous  Table,    . 

.  40 

French  Measures  and  Weights, 

40 

IV.  ADDITION. 

Definitions,  Illustrations,  and  Expli 

nations, 41 

Simple  Addition,  ....  42 

Compound  Addition,    ...  44 

Compound  and  Simple  Addition  com 

pared, 45 

Methods  of  Proof,         .        .        .        .  4e 
Importance  of  Accuracy  and  Certainty,  4S 

Problems, 5C 

Shorter  Method  of  adding  Compound 

Numbers, 54 

Common  Metliod,  .        .        .        .55 

Examples  for  Practice, .        .        .        .56 
Addition  of  several  Columns,       .        .  58 

Leger  Columns, 59 

(9\ 


CONTENTS. 


V.   SUBTRACTION. 

Definitions  and  Illustrations,        .        .  61 
Method  when  Problems  require  no  Re- 
duction,       61 

Methods  of  Proof,         .        .        .        .62 

Problems  requiring  no  Reduction,        .  63 
Bimple  Subtraction,      .       .        .        .64 

Compound  Subtraction,        .        .        .66 
Problems,       .        .        .        .        ...  67 

The  changed  Minuend  not  written,    .  68 
Subtrahend  Figure  may  be  increased,  69 
Shorter  Method  of  subtracting  Com- 
pound Numbers,        ...  70 
Problems,       ...                .        .  71 

Subtraction  from  Left  to  Right,    .        .  73 
Subtraction  of    several    Numbers   at 
once, 74 

VI.   MULTIPLICATION. 

Definitions  and  Illustrations,  .  .  76 
Product  not  affected  by  Change  in  Order 

of  Factors, .        .  ...  78 

Simple  Multiplication — when  only  one 

Factor  is  a  large  Number,  .  .  79 
Compound  Multiplication,    .        ,        .80 

Methods  of  Pr«of, 81 

Problems.  Multiplier  a  single  Figure,  82 
Reduction  Descending,  .  .  .85 
Multiplication  by  Factors,  .  .  .88 
Both  Factors  large  Numbers,  .  .  90 
Abbreviated  Method 93 

VIL  DIVISION. 
Definitions  and  Illustrations,  .  .  97 
Method-^  of  Proof,  .  .  .  .100 
Examples.  Cluotient  a  single  Figure,  101 
Dividend  a  larjre  Number,  .  .  .103 
Decimal  Fractions  in  the  Quotient,  .  104 
Compound  Division,  ....  ]05 
Problems  for  Solution,  .  .  .106 
Reduction  Ascending,  ,  .  .  107 
Division  by  Factors,  .  .  .  .110 
Divisor  a  Jarge  Number,  .        .112 

Lon<;  Divisit  n, 116 

Abbreviated  Process,  .  .        .117 


VIII.    CONTRACTIONS     AND     MIS- 
CELLANEOUS PROBLEMS. 

To  multiply  by  5,  12J,  16f,  &c.,  .  119 

One  Part  of  Multiplier  a  Factor  of 

anothtr  Part, 120 

To  divide  b>  99,  999,  &c.,  .  .121 

To  divide  by  5,  16 j,  25,  ice,       .        .  122 


Miscellaneous  Problems,    .       .  1,23 

Bills  of  Goods, 126 

Examples  for  Practice,       .  130 

IX.     PROPERTIES     OF     NUMBERS, 

DIVISORS,   MULTIPLKS,  &c. 

Definitions,  ...  .133 

Demonstration  of  Principles,  134 

Divisibility  by  2,  5,  3J,  &c.,  135 

''  "    2,  25,  1C|,  &c.,    .  135 

"  «   8,  125, 333J,  &c.,        .  136 

"  "9 136 

'•  "3,         ....  139 

"  "11,        .        .        .        .  139 

Other  Tests, 141 

Recapitulation,  .  .  .  .  .141 
Definitions  of  Factors,  Powers,  &.C., .  142 
Method  of  Factoring,  .        .        .        .143 

Exercises, 145 

Common  Divisor,  Definitions,  Sec,  .  140 
Greatest  Common  Divisor  by  Factors,  146 
A  more  brief  Method, .  .  .  .147 
Factoring  not  necessary,      .  .  148 

Method  by  Addition  and  Subtractlor,  149 
Demonstration  of  Method  by  Divisio.\,  150 
Application  of  Method  by  Divisii.r.,  150 
Common  Multiples,  Definitions,  &.t  ,  15? 
Least  Common  Multiple  by  Factci..  15^ 
Abbreviated  Method,  .        .  ISO 

When  Factors  cannot  easily  be  found    15ft 

X.   FRACTIONS. 

Introductory, IJV 

Definitions  of  Halves,  Thirds,  tc,  .  la. 
Fractional  Parts, .  .  .  \A 
Method  of  writing  tractions,  .  .  l^t 
Explanation  of  Fractitms,  .  .  .  I<i0 
Various  Kinds  of  Fractions.  .  .  161 
Operations  on  Fractions  illustrated,  .  16i 
Reduction  to  Improper  Fractions,  .  16J 
Reduction  to  Whole  or  Mixed  Num- 
bers,   163 

Miscellaneous  Operations, .  .  .164 
One    Number    a    Fractional   Part  of 

another,    .  ...  167 

Other  Methods  of  expressing  the  Value 

of  a  Fraction,  .  ...  170 

To  find  a  Fractional  Part  of  a  Number,  170 
To  multiply  by  a  Fraction,  .  .  179 
Practical  Problems,  .  .  .  .174 
Multiplication   and  Division  of    the 

Numerator, 178 

Multiplication  of  the  Denominator,  .  179 
Division  of  the  Denominator,     .       .  179 

• 


CONTENTa. 


XI 


Recapitulation  and  Inferences,   .        .  180 
Multiplication  and   Division  of  both 
Numerator  and  Denominator  by  the 
same  Number,  .        .        .        .180 

Lowest  Terms, 181 

Cancellation, 163 

Compound  Fractions, ....  186 
Second  Method,  .        .        .        .187 

Multiplication  of  Fractions,  .  .  188 
Reduction  of  a  Vulgar  to  a  Decimal 

Fraction, 190 

Fractional  Part  of  Denominate  Num- 
bers, ....'...  192 
To  find  a  Number  from  its  Fractional 

Part, 196 

Practical  Problems,  .  .  .  .197 
Division  by  Fractions,  .  .  -.  200 
Process  of  Division  generalized,  .  203 
Complex  Fractions,  ....  205 
Otber  Changes  in  the  Terms  of  a  Frac- 
tion,   207 

Common  Denominator,  .  .  .  908 
Addition  and  Subtraction  of  Fractions,  210 

XI.  APPUCATIONS  OF  FOREGOING 

PRINCIPLES. 
Introductory  Note,       ....  213 
Miscellaneous  Problems,     .        .        .  214 
Practice, 218 

XIL   RATIO   AND   PROPORTION. 

Definitions  and  Illustrations  of  Ratio,  222 
Reduction  of  Ratio,             .  .  223 
Definitions  and  Illustrations  of  Pro- 
portion,      225 

To  find  a  missing  Term,     .  .  226 

Relations  of  Terms,  ....  228 
Practical  Problems,  ....  228 
Problems  in  Compound  Proportion,    .  230 

XIIL   DUODECIMAL   FRACTIONS. 
Definitions,  ....  .  2.33 

Problems,     .        .       .        .        .        .234 

XIV.  ALUGATION. 
Definitions  and  Explanations,    .        .  237 
Problems, 237 

XV.  INTEREST   AND   BUSINESS 
PROBLEMS. 

Introductory, 210 

Definitions, 241 

Legal  Rate,  .  ....  343 


Interest  for  2  Months,  200  Months, 

&c.,  at  6  per  cent.,  .... 
Recapitulation  and  Inferences,  . 
Table  showing  Interest  for  convenient 

Fractional  Parts  of  200  Months,  20 

Months,  &.C.,  at  6  per  cent.,     . 
Questions  on  the  Table, 
Applications  of  the  Table,  . 
Interest  for  various  Times, 
Computation  of  Time, 
Interest  by  Days,        .... 
Interest  by  Dollars  for  Months,  . 
Interest  by  Dollars  for  Days, 
When  to  disregard  Cents,   . 
Interest   at    various    Rates   obtained 

from  that  at  6  per  cent.. 
Interest  at  various    Rates    obtained 

directly, 

Promissory  Notes, 

Negotiable  Notes,  Indorsements,  and 

Protests, 

Joint  and  Several  Notes, 

Renewal  of  Notes,       .... 

Banks  and  Banking,    .... 

English  Method  of  computing  Interest, 

Partial  Payments,        .        .        .        . 

Merchants'  Method,    .        .        .        . 

To  find  the  Time,        .... 

Equation  of  Payments, 

To  find  Date  nf  Equated  Time, . 

Equation  of  Accounts, 

To  find  the  Principal  or  Interest  from 

the  Amount, 

Discount  and  Present  Worth, 

Business  Method  of  Discount,    . 

To  find  the  Rate,  .... 

To  find  the  Principal  from  the  Interest, 
Compound  Interest,     . 
Table  for  Compound  Interest,    . 
Commission,         .  . 

Stocks,.        .        .  .        .        . 

Insurance,    . 

Assessment  of  Taxes, 

Orders, . 

Bills  of  Exchange, 

Profit  and  Loss,   . 

Partnership, 

Partnership  on  Time, 


243 
245 


246 
247 
247 


255 
256 
257 


259 


264 

266 
369 
270 
272 
275 
277 
281 
283 
284 
287 
290 


293 
296 
298 
300 
301 
302 
303 
334 
307 
309 
310 
311 
313 
316 
333 
334 


XVI.   POWERS   AND  ROOTS. 

Definitions, 327 

Denominations  of  Squares  and  Roots,  328 
Division  info  Periods,  ....  329 
Method  of  forming  a  Square,  .  339 


zu 


CONTENTS. 


Method  of  extracting  the  Square  Root,  331 
Rule  for  Square  Root,  with  Problems,  332 
Square  Root  of  Fractions,  .  .  .  333 
Square  Root  of  Decimal  Fractions,  .  334 
Cube  Root.  —  Relation    of  Cube  to 

Root, 

Division  into  Periods, . 
Method  of  forming  a  Cuhe, 
To  extract  the  Cube  Root,  . 
Rule  for  the  Cube  Root,     . 
Cube  Root  of  Fi  actions,     . 
Cube  Root  of  Decimal  Fractions 
Rule  for  extracting  a  Root  of  any  De- 
gree,  


341 


XVI r.   MENSURATION. 
Polygons,      .  ....  342 

Problems,  344 

Propeitics  of  the  Right-angled  Trian- 
gle, with  Problems, .        .        .        .345 


Solids,  . 
Problems, 


34fl 
348 


XVIII.   PROGRESSIONS. 

Arithmetical  Progression,  .        .        .  349 
Problems,     .        .  .  350 

To  find  the  Sum  of  a  Series,  .  350 

Problems,      ...  .351 

Geometrical  Progression,    .        .        .351 

Problems, 352 

To  find  the  Sum  of  a  Geometrical 
Series,      ......  352 

Problems, 353 

Infinite  Decreasing  Series, .        .        .354 

XIX.  CIRCULATING  DECIMALS,  354 

XX.  MISCELLANEOUS    EXAM- 
PLES,      .       .  SM 


AEITHMETIC. 


SECTION,  ■t,;,;,_. ..!.!.  _.. 

1 .    Definitions, 

1.  Quantity  is  a  term  applied  to  whatever  may  be  in- 
creased, diminished,  or  measured. 

2.  A  CONCRETE  UNIT  is  any  quantity  which  may  be  con- 
wdered  by  itself,  and  made  the  measure  of  other  similar 
quantities. 

(a.)  It  may  be  a  single  thing,  as  an  apple,  a  book,  a  pound,  a  foot,  a 
dolhir ;  or  it  may  be  a  collection  of  single  things,  as  a  dozen  of  apples^ 
a  score  of  sheep. 

3.  An  ABSTRACT  UNIT  is  the  idea  or  conception  of  one,  or 
of  unity,  without  reference  to  any  particular  thing  or  quan- 
tity. Any  number  of  abstract  units,  considered  as  forming  a 
single  collection  or  whole,  may  also  be  regarded  as  an  ab- 
stract unit ;  as,  one  ten,  one  hundred,  one  thousand,  &c. 

(a.)  The  term  unit,  when  used  without  any  limitation,  always  refers 
to  a  single  thing  or  quantity,  or  to  the  abstraction,  one. 

4.  A  CONCRETE  NUMBER  exprcsscs  a  single  thing  or 
quantity,  considered  as  a  unit  of  measure  ;  or  it  expresses  how 
many  such  units  there  are  in  any  given  quantity ;  as,  one 
yard,  three  shillings,  eight  barrels,  ten  birds. 

5.  An  ABSTRACT  NUMBER  cxprcsscs  how  many  times  a 
unit  is  taken  or  repeated,  without  any  reference  to  the  nature 
of  the  unit ;  as,  one,  eight,  twelve. 

6.  The  unit  which  any  number  expresses  is   called  the 

UNIT  OF  MEASURE,  Or  the  UNIT  OF  COMPARISON. 

{a.)  In  an  abstract  number  it  is  the  abstract  unit,  and  in  a  concrete 
1  (1) 


3  DEFINITIONS. 

number  it  is  the  unit  which  the  number  measures.  Thus,  in  seven  dollars 
it  is  a  dollar,  in  twelve  bushels  it  is  a  bushd,  in  forty-three  dozen  of  eggs  it  is 
a  dozen  of  eggs,  in  six  tens  it  is  one  ten,  &c. 

7.   Arithmetic  is  the  science  of  numbers  and  art  op 

NUMERICAL    COMPUTATION. 

(a.)  It  treats  of  numbers  with  reference  to  their  nature  and  use,  their 
properties  and  relations  ;  explains  the  various  methods  of  representing 
them  ;  and  .:iri'(^ludes  ih^  thticrj  of ;  all  numerical  operations,  as  well  as 
the  practical  ihethods  ©f  pepfoKming  them. 

8.,\\l^'be.  f?pe?atii;)n^  of  ^vbich  ryumbers  are  susceptible  are 
four  in  hurnber ;  viz.,  A'ddition^  Subtraction,  Multiplication^ 
and  Division. 

9.  Among  the  characters  used  to  indicate  numerical  opera- 
tions or  relations  are  the  following :  — 

(a.)  =:  The  sign  of  equality,  called  equal,  or  equal  to,  sig- 
nifies that  the  quantities  between  which  it  is  placed  are  equal 
to  each  other. 

(Jb.)  -\-  The  sign  of  addition,  called  plus  or  and,  signifies 
that  the  quantities  between  which  it  is  placed  are  to  be  added 
together. 

Thus,  7  4"  4  =  11,  is  read,  seven  plus  four  equal  eleven  ;  or,  seven  and 
four  equal  eleven  ;  and  means  that  seven  added  to  four  equal  eleven. 

(c.)  —  The  sign  of  subtraction,  called  minus  or  less,  Signi- 
fies that  the  number  following  it  is  to  be  subtracted. 

Illustration.  9  —  6  =  3,  is  read,  nine  minus  six,  or  nine  less  six,  equal 
three,  and  means  that  nine  diminished  by  six  equal  three. 

(d.)  X  The  sign  of  multiplication,  called  times  or  multi- 
plied by,  signifies  that  the  numbers  between  which  it  is  placed 
are  to  be  multiplied  together. 

Illustration.  7  X  5  =  35,  is  read,  seven  times  five  equal  thirty-Jive ;  or, 
seven  multiplied  by  five  equal  thirty-five. 

(e.)  -7-  The  sign  of  division,  called  divided  hy,  means  that 
the  number  before  it  is  to  be  divided  by  that  following  it. 

Illustration.  12-5-3  =  4,  is  read,  twelve  divided  by  three  equal  four, 
and  means  that  twelve  divided  by  three  gives  four  for  a  quotient. 

(/.)    Division  may  also  be  expressed  by  writing  the  num- 


NOTATION    AND    NUMERATION.  3 

ber  to  be  divided  above  the  number  hj  which  it  is  tc  be 
divided,  with  a  line  between  them. 

12 

Illustration.      —  =  4,  means  the  same  as  12 -5- 3  =4  j  i.  e.,  that  the 

quotient  of  twelve  divided  by  three  equals  four. 

(«/.)  Such  expressions  as  J^^  J^s.^  |.j  are  usually  called  fractions,  and 
read  thus:  twelve  thirds,  sixteen  fourths,  seven  eighths ;  though  they  may, 
with  equal  correctness,  be  read  as  twelve  divided  by  three,  sixteen  divided 
by  four,  serpen  divided  by  eight.  When  read  as  fractions,  the  number  above 
the  line  is  called  the  numerator,  and  the  number  below  it  the  denominator. 
See  Section  X. 

(A.)  The  more  common  use  of  fractions  is  to  express  the  value  of  one  or 
more  such  parts  as  are  obtained  by  dividing  a  unit  into  a  given  number  of 
equal  parts,  or,  which  is  the  same  thing,  to  express  the  value  of  one  or  more 
equal  parts  of  such  kind  that  a  given  number  of  them  will  equal  a  unit. 
Thus,  3  (read  three  fourths)  is  used  to  express  the  value  of  three  such 
parts  as  would  be  obtained  by  dividing  a  unit  into  four  equal  parts ; 
or,  in  other  words,  the  value  of  three  equal  parts  of  such  kind  that  it 
would  take  four  of  them  to  equal  a  unit. 

(i.)  The  numerical  value  of  a  fraction  is  the  same,  whether  we  con- 
sider that  it  expresses  a  division  to  be  performed,  or  a  certain  number 
of  equal  parts,  and,  in  either  case,  it  is  obvious  that  a  fraction  must  equal 
unity  whenever  its  numerator  equals  its  denominator.     Thus,  l  =  a  = 


&c. 


SECTION   II. 

NOTATION  AND  NUMERATION. 

2,    Definition  of  Terms. 

Notation  and  Numeration  treat  of  the  various  methods 
»f  representing  and  expressing  numbers. 

(a.)  The  distinction  usually  made  between  notation  and  numeration 
is,  that  the  former  treats  of  the  methods  of  representing  numbers  by 
written  characters,  while  the  latter  treats  of  the  methods  of  reading 
tijem,  or  of  expressing  them  in  words. 


4  NOTATION    AND    NUMERATION. 

3.    Methods  of  representing  Numbers,  and  Origin  of  the 
Decimal  System. 

Numbers  may  be  represented  by  material  objects  or  visible 
marks,  by  words,  and  by  figures. 

(a.)  As  our  ideas  of  number  are  derived  primarily  from  material  ob- 
jects, so  the  most  natural  and  obvious  method  of  communicating  tliem 
to  others  is  by  exhibiting  as  many  such  objects  as  there  are  units  in  the 
number  considered. 

(6.)  It  is  probable  that  in  the  earlier  stages  of  society  numbers  were 
represented  only  in  this  way,  the  fingers  being,  as  a  general  thing,  made 
use  of  as  counters.  Thus,  three  fingers  would  be  shown  as  a  symbol  for 
the  number  three,  five  fingers  for  the  number  five,  and  the  fingers  of 
both  hands  for  the  number  ten. 

(c.)  Such  a  method  would  naturally  lead  a  people  using  it  to  repre- 
sent large  numbers  by  exhibiting  the  fingers  of  both  hands  as  many 
times  as  there  are  tens  in  the  numbers  considered,  and  by  thus  leading 
them  to  reckon  by  tens,  would  lay  the  foundation  for  a  system  of  num- 
bers similar  to  the  one  in  general  use,  which  is  known  as  the  decimal* 

SYSTEM. 

4:,    Nature  of  the  Decimal  System  of  Numbers. 

{a.)  The  fundamental  idea  of  the  Decimal  System  is,  that  ten  single 
things  may  be  regarded  as  forming  a  single  collection  or  group  :  ten  of 
these  groups  as  forming  a  larger  group ;  and  so  on,  ten  groups  of  one 
size  forming  a  new  group  of  a  larger  size,  each  capable  of  being  regarded 
and  dealt  with  as  a  single  thing  or  unit.  This  idea  renders  it  easy  to 
represent  the  largest  numbers,  by  having  names  for  each  of  the  first  ten 
numbers,  and  for  each  group  formed  by  combining  ten  of  the  smaller 
ones. 

{h.)  In  conformity  with  it,  we  might  count  thus :  one,  two,  three,  four^ 
Jive,  six,  seven,  eight,  nine,  ten,  one  and  ten,  two  and  ten,  three  and  ten,  four 
and  ten.  Jive  and  ten,  six  and  ten,  seven  and  ten,  eight  and  ten,  nine  and  ten, 
two  tens,  tiro  tens  and  one,  two  tens  and  two,  &c.,  to  nine  tens  and  eight, 
nine  tens  and  nine,  ten  tens  or  one  hundred,  one  hundred  and  one,  &c.,  to  nine 
hundreds  nine  tens  and  eight,  nine  hundreds  nine  tejis  and  nine,  ten  hundreds 
or  one  thousand,  &c. 

(c.)  Adopting  this  method,  and  forming  compound  words  by  drop- 
ping the  conjunction,  we  should  count  from  ten  thus :  one-ten,  two-ten, 
three-ten,  four -ten.  Jive-ten,  six-ten,  seven-ten,  eight-ten,  nine-ten,  two-tens,  two- 
tenii-one,  two-tens  two,  &c. 

*  The  word  decimal  is  derived  from  the  Latin  word  decern,  which  signi- 
fleB  ten. 


NOTATION    AND    NUMERATION.  5 

(d.)  Changing  the  word  ten  to  teen,  and  dropping  the  hyphen  in 
counting  from  ten  to  two-tens,  we  should  have  oneteen,  twoteen,  threeteen^ 
fourteen.  Jiveteen,  sixteen^  seventeen^  eighteen,  and  nineteen. 

(e.)  By  now  changing  Jive  to  fif,  three  to  thir,  and  substituting  for 
one-teen  and  two-teen  the  words  eleven  and  twelve,  signifying  respectively 
one  left  and  two  left,  (i.  e.,  ten  and  one  left,  ten  and  two  left,)  we  should 
have  the  familiar  names,  eleven,  twelve,  thirteen^  fourteen,  fifteen,  sixteen^ 
seventeen,  eighteen,  and  nineteen. 

(f)  Substituting  the  syllable  ti/  for  tens  in  the  words  two-tens,  three- 
tens,  &c.,  would  give  the  words  twoty,  threety,  fourty,  fivety,  sixty,  seventy, 
eighty,  and  ninety;  and  again  changing  the  thi-ee  to  thir,four  to  for,  and  the 
^ve  to  ff  and  substituting  twen,  derived  from  twain,  for  two,  would  give 
the  familiar  names  twenty,  thirty,  forty,  fifty,  sixty,  seventy,  eighty,  and 
ninety. 

^g.)  These  changes  would  enable  us  to  count  from  two-tens  or  twenty 
thus :  twenty-one,  twenty-two,  &c.,  to  twenty-nine,  thirty,  thirty-one,  &c.,  to 
me  hundred,  one  hundred  and  one,  &c. 

(A.)   This  method  of  expressing  numbers  is  the  one  now  in  general  use. 

5,    Arabic  and  Roman  Methods  of  Notation. 

(a.)  Numbers  are  usually  represented  to  the  eye  by  charac- 
ters called  figures,  though  sometimes  by  letters  of  the  alphabet. 

(b.)  The  method  by  figures  is  called  the  Arabic  Method, 
because  it  was  introduced  into  Europe  by  the  Arabs. 

Note.  —  The  Arabs  probably  obtained  it  from  the  Persians,  who  had 
obtained  it  from  the  Hindoos.  Its  origin  has  never  been  satisfactorily 
determined. 

(c.)  The  method  by  letters  is  called  the  Roman  Method^ 
b  ^cause  it  was  used  by  the  ancient  Romans. 

6.    The  Figures. 
The  figures  ai-e  ten  in  number,  viz. :  — 

1  or  /^  called  one. 

2  or  S ,  called  two. 

3  or  3 ^  called  three. 

4  or  ^^  called  four. 

5  or  e5^    called  jive. 


6  or  6 )  called  six. 

7  or  y J  called  seven, 

8  or  8 ,  called  eight. 

9  or  ^ ^  called  nine. 

0  or  0 ^  called  zero,  ciphery 
nothing,  or  nought 


«  NOTATION    AND    NTMERATION. 

y.    The  Place  of  a  Figure  determines  its  Denominatim*. 

Each  of  these  figures  represents  as  many  units  as  its  nam* 
indicates ;  but  the  size  or  denomination  of  those  units  is  deter- 
mined by  the  place  or  position  of  the  figure  with  refecence  to 
a  period  or  dot,  called  the  decimal  point, 

8.    Names  and  Position  of  the  Decimal  Places. 

(a.)  The  figure  immediately  at  the  left  of  the  point  repre- 
sents ones,  or  simple  units ;  the  second  figure  at  the  left  repre- 
sents tens,  (i.  e.,  units  of  the  denomination  or  value  of  ten 
ones,  or  ten  simple  units ;)  the  third  figure  represents  hun- 
dreds; the  fourth  represents  thousands,  and  so  on;  the  figure 
in  any  place  always  representing  ten  times  the  value  it  would 
represent  if  it  stood  one  place  farther  towards  the  right. 

(h.)  Hence  each  place  has  its  peculiar  name,  the  first, 
second,  third,  and  fourth  places  being  called,  respectively,  the 
units'  place,  the  tens'  place,  the  hundreds'  place,  and  the  thou- 
sands' place.  Moreover,  the  position  of  these  places  is  marked 
by  the  figures  occupying  them.  Hence  each  figure  performs 
a  double  office,  viz.,  it  marks  a  place,  and  indicates  as  many 
of  the  denomination  of  that  place  as  its  name  indicates. 

(c.)    The  following  will  illustrate  this  :  — 


o   a  H  -^   a 

0000- 

(d.)  In  the  above  example  each  zero  marks  a  place,  and 
shows  that  there  are  none  of  the  denomination  of  tlie  place  it 
occupies  expressed  in  that  place. 

(e.)  As  another  illustration,  take  the  expression  2503. 
Here  each  figure  marks  a  place,  and  denotes  as  many  of  the 
denomination  of  that  place  as  its  name?  implies;  i.  e.,  the  3 
marks  the  units'  place,  and  shows  that  there  are  3  units ;  the 
0  marks  the  tens'  place,  and  sliows  that  there  are  no  tens ;  the 
5  marks  the  hundreds'  place,  and  shows  that  there  are  5  hun- 
dreds ;  and  the  2  marks  the  thousands'  place,  and  shows  that 
there  are  2  thousands.  The  number  is  read  two  thousand  fivt 
hundred  and  three. 


NOTATION    AND    NUMILRATION.  7 

Note.  —  Tlie  zero  is  often  called  an  insignificant  figure,  and  the  other 
nine  digits  significant  figures ;  but  there  is  no  foundation  for  the  distinc- 
tion. The  zero  performs  an  office  precisely  similar  to  that  of  any  other 
figure,  as  the  above  explanation  of  the  use  of  the  figures  used  in  writing 
2503  clearly  shows.  Even  when  standing  by  itself  it  is  as  expressive  as 
any  other  figure. 

{/.)  The  decimal  point  is  often  omitted  in  writing  numbers ; 
but  in  all  such  cases  it  is  understood  to  belong  at  the  right  of 
the  given  number,  thus  making  the  right  hand  figure  represent 
units. 

O.    Method  of  reading  Numbers. 

In  reading  numbers  expressed  by  figures  we  begin  at  the 
left  hand,  i.  e.,  with  the  highest  denomination. 

{a.)  546  =  five  hundreds,  four  tens,  and  six  units,  and  is  read  Jive 
hundred  and  forty-six. 

(6.)  398  =  three  hundreds,  nine  tens,  and  eight  units,  and  is  read  three 
hundred  and  ninety-eight. 

(c.)  407=  four  hundreds,  no  tens,  and  seven  units,  and  is  read  four 
hundred  and  seven. 

[d.)  180=  one  hundred,  eight  tens,  and  no  units,  and  is  read  one 
hundred  and  eighty.  ^ 

(e.)    64  or  064  =  six  tens  and  four  units,  and  is  read  sixty  four. 

1 0 .    Exercises  in  reading  and  writing  Numbers. 

Read  the  following  numbers,  and  also  give  the  value  of 
each  in  hundreds,  tens,  and  units :  — 

7.  031. 

8.  37. 

9.  200. 

]  0.  Explain  the  use  of  each  figure  used  in  the  above  num- 
bers, as  in  the  following  model :  — 

Model.  —  In  the  first  number,  507,  the  7  marks  the  units'  place,  and 
shows  that  there  are  7  units  ;  the  0  marks  the  tens'  place,  and  shows  that 
there  are  0  tens ;  the  5  marks  the  hundreds'  place,  and  shows  that  there 
are  5  hundreds. 

11.    How  will  you  write  four  hundred  and  seven? 

Ans.  By  writing  4  in  the  hundreds'  place,  0  in  the  tens',  and  7  in  the 
i^nits' ;  tl]us,  407. 


1.   507. 

4.    379. 

2.    409. 

0.    211. 

3.    5-28. 

6.    403. 

8  NOTATION    AND    NUMERATION. 

12.  How  will  you  write  two  hundred  and  seventeen? 

13.  How  will  you  wTite  eight  hundred  and  forty-one  ? 

14.  How  will  you  write  eight  hundred  and  twelve  ? 

15.  How  will  you  write  seven  hundred  and  forty-six? 

1 6.  How  will  you  write  six  hundred  and  ninety-four  ? 

17.  How  will  you  write  nine  hundred  and  sixty-four? 

18.  How  will  you  write  four  hundred  and  sixty-nine? 

1 9.  How  will  you  write  nine  hundred  ? 

20.  How  will  you  write  seven  hundred  and  eighty  ? 

11.  Number  of  Decimal  Places  unlimited. 
Extending  these  principles,  we  can  take  as  many  places  a3 
we  please,  by  giving  to  the  figure  in  each  ten  times  the  value 
it  would  have  if  written  one  place  farther  to  the  right.  The 
names  of  the  places  as  far  as  the  twenty-fourth  are  given  in 
the  following  example. 

'-       i       I       ^       ^       ^       i 

.2  S    •         3  "         fi  i  a  S 

?i^  &5i  721  ?l  .  2S     11     nli  i 

^^i  ^o<«     -So-^     -5^:2  -SSg  ^So  ^£§  "S^i^ 

CctJ  Cc-5     fie*     5c~  cAa  Ces  OqO  Sc--S 

3S>1  sga     sss     SaJ-E  =!§:=  s§3  sg^a  flScS 

tSrHtn  KEHC?«HC?a3HH  WHM  KhS  «hH  PaH^sp, 

000,00  0,000,000,000,000,000,000. 

"  "  g  8  s 


s§§  §ij  s§g  s§§  s§s  sgs  gg^ 

5.0,0.     a-aa  0.0.0.  ftaa  1 1^  ^    ^  |  -2  §|  i 

j3  ;a     xi  Si  Si  Si  Si  a  Ji  Si  Si  «•  «-  *     &  =•  *  a  o.  «• 

§1    g«HS  is2  SSS  tit    t%%  S552 


£^:a-s£5    ^-'^^    ■??-^-^    ^^-^     -"-     ---     *«^ 
sis    53§  2 

IS.  Division  into  Periods, 
(a.)  By  inspecting  the  above  example  it  will  be  seen  that  the 
first  three  places  are  occupied  by  units,  tens,  and  hundreds,  — 
the  second  three  by  thousands,  tens  of  thousands,  and  hundreds 
of  thousands,  —  the  third  three  by  millions,  tens  of  millions, 
and  hundreds  of  millions,  and  so  on.  If  the  first  three  places 
were  called,  as  they  might  be  with  perfect  propriety,  units, 
tens  of  units,  and  hundreds  of  units,  and  we  should  divide  the 
number  into  periods  of  three  figures  each  by  commas,  the  first 
period  would  be  the  period  of  units,  the  second  the  period  of 
thousands,  the  third  of  millions,  the  fourth  of  billions,  &c. 


NOTATION    AND    NU:\[KIlATION. 


9 


{L)  The  right  hand  figure  in  each  period  expresses  units 
or  ones  of  the  denomination  of  that  period,  while  the  second 
figure  expresses  tens,  and  the  third,  or  left  hand  figure,  ex- 
presses hundreds  of  that  denomination. 

(c.)    This  is  exhibited  in  the  following  table :  — 
*s  i. 


o 

000 


<y^ 


EH-g 


m-n 


H^ 


t3 

2      I 


000,000,000,000,000,0  00,000 


E'„  S 


.     ft    .    « 


G.    .2—4'     .2  ~   ** 


m    .2    . 


-e  -a  -2     73:2 
2   3=      £  ? 


•2    c 

o  o 


■Si     ^  ^  J= 


O    O    O        o    o    o 

^  £  £    j=  ji  js 


•a  s  m  -a  a  a 
«*  :3  a  a)  o  eJ 
t;   -  o     i  j:  2 


III    ^|5 

o   o  o      So© 


.2   a 
«   3 


3    55 


l'>        CO   >o 


13.    Exercises  to  secure  Familiarity  with  the  Places  and 

^  Periods. 

1.  What  is  the  name  of  the  first  period  at  the  left  of  the 
point?  of  the  second  period  ?  of  the  third?  of  the  fourth?  of 
the  fifth  ?  of  the  sixth  ?  of  the  seventh  ?  of  the  eighth  ? 

2.  What  is  the  name  of  the  period  occupying  the  first,  sec- 
ond, and  third  places  at  the  left  of  the  point  ? 

3.  Occupying  the  fourth,  fifth,  and  sixth  places  ? 

4.  Occupying  the  seventh,  eighth,  and  ninth  ? 

5.  Occupying  the  tenth,  eleventh,  and  twelfth  ? 

6.  Occupying  the  thirteenth,  fourteenth,  and  fifteenth  ? 

7.  Occupying  the  sixteenth,  seventeenth,  and  eighteenth  ? 

8.  Occupying  the  nineteenth,  twentieth,  and  twenty-first  ? 

9.  Occupying  the  twenty-second,  twenty-third,  and  twenty 
fourth  ? 

10.  Wliat  is  the  number  of  the  millions'  period  ? 
Ans.  The  third  period  at  the  left  of  the  i)oint. 


10 


NOTATION    AN1>    NffMERATION. 


11.  What  is  the  number  of  the  thousands'  period  ? 

12.  What  is  the  number  of  the  quintillions'  period  ? 

13.  What  is  the  number  of  the  trillions'  period  ? 

14.  What  is  the  number  of  the  units'  period? 

15.  What  is  the  number  of  the  sextillions'  period? 

1 6.  Wliat  is  the  number  of  the  billions'  period  ? 

17.  "What  is  the  number  of  the  quadrillions'  period  ? 

How  many  places  are  there  between  the  point  and  the 


22.  Thousands'  period  ? 

23.  Billions' period? 

24.  Sextillions'  period  ? 

25.  Trillions'  period  ? 


18.  Millions' period  ? 

19.  Quintillions' period  ? 

20.  Units'  period  ? 

21.  Quadrillions' period  ? 

26.  In  which  place  of  what  period  would  the  fourth  figure 

at  the  left  of  the  point  be  ? 

Ans.   In  the  first  place  of  the  second  or  thousands'  period. 

27.  In  which  place  of  what  period  would  the  seventh  fig- 
ure at  the  left  of  the  point  be  ? 


28.  Would  the  tenth  ? 

29.  Would  the  sixteenth  ? 

30.  Would  the  second  ? 

31.  Would  the  eleventh  ? 

32.  Would  the  twentieth  ? 

33.  Would  the  twenty-second  ? 

34.  Would  the  third  ? 

35.  Would  the  twelfth  ? 

36.  Would  the  twenty-first? 

37.  Would  the  sixth  ? 


38.  Would  the  thirteenth  ? 

39.  Would  the  seventeenth  ? 

40.  Would  the  fifth? 

41.  Would  the  fourteenth  ? 

42.  Would  the  twenty-third  ? 

43.  Would  the  eighth  ? 

44.  Would  the  seventeenth  ? 

45.  Would  the  twenty-fourth  ? 

46.  Would  the  eighteenth  ? 

47.  Would  the  fifteenth  ? 


48.  What  would  be  the  denomination  of  a  figure  in  each 
of  the  above-named  places  ? 

Ans.  The  denomination  of  a  figure  in  the  fourth  place  at  the  left  of 
the  point  would  be  thousands,  that  of  a  figure  in  the  seventh  place  at 
the  left  would  be  millions,  that  of  the  tenth  would  be  billions,  &c. 

49.  Where  must  a  figure  be  placed  to  represent  trillions  ? 
Ans   In  tho  first  place  of  the  fifth  period,  which  is  tlie  thirteenth  place, 

at  tlie  left  of  the  point. 


NOTATION    AND    NUMERATION. 


11 


Where  must  a  figure  be  placed  to  represent 


50.  Millions? 

51.  Ten-minions  ? 

52.  Units? 

53.  Ten-thousands? 

54.  Quadrillions  ? 

14. 


55.  Thousands  ? 

6Q.  Hundred-trillions? 

57.  Hundreds? 

58.  Hundred-billions? 

59.  Ten  biUions  ? 


Method  of  reading  lumbers. 
Exercises. 

(a.)  In  reading  a  number  represented  by  figures,  we  ordina- 
rily commence  at  the  left  hand,  and  read  each  period  as  though 
its  figures  stood  alone,  giving  afterwards  the  name  of  the  period. 

For  instance,  the  number  42,000,070,294,600,706  would  be  read  in  the 
same  way,  and  would  express,the  same  value,  as  if  written  42  quadrillions, 
70  billions,  294  millions,  600  thousands,  and  706. 

Note.  —  The  scholar  should  regard  a  mistake  in  reading  numbers  aa 
one  of  the  most  dangerous  which  can  be  made,  for  lie  will  not  only  be 
likely  to  copy  the  numbers  in  the  same  manner  as  he  reads  them,  but  he 
will  give  those  to  whom  he  reads  a  false  idea,  which,  unless  they  have 
the  figures  before  them,  they  cannot  correct. 

(b.)    Read  each  of  the  following;  numbers  :  — 


1. 

43,271. 

16. 

607,429. 

2. 

500,207. 

17. 

1,579,432. 

3. 

24,000,217. 

18. 

914,307,426. 

4. 

53,279,412. 

19. 

53,729,415. 

5. 

432,160,023. 

20. 

21,437,986,512. 

6. 

70,000,000. 

21. 

42,000,042,042. 

7. 

86,102,102. 

22. 

547,547,547,547. 

8. 

150,437,986,216. 

23. 

101,101,101,101. 

9. 

20,020,020,020. 

24. 

2,002,002,002,002. 

10. 

200,200,200,200. 

25. 

130,201,040,999,999. 

11. 

70,000,007,700,077. 

26. 

73,006,200,474. 

12. 

2,008,002,008,002,008. 

27. 

900,000,726,000. 

13. 

6,000,070,000,600,007. 

28. 

74,206,372,704. 

14. 

3,200,020,006,307,004. 

29. 

407,000,000,030,002. 

15. 

73,052,700,060,007. 

30. 

703,700,000,000,006. 

(c.)    Explain  the  use  of  the  figures  used  in  writing  the  above 
numbers.     (See  model  following  the  10th  example  in  10.) 


12  NOTATION    AND    NUMERATION. 

1^.    All  intervening  Places  to  he  jiUed. 

(a.)  When,  as  is  usually  the  case,  the  places  are  only  marked 
by  the  figures  occupying  them,  every  place  between  any  given 
figure  and  the  point  must  be  occupied  by  some  one  of  the 
digits,  for  otherwise  it  will  be  diflBcult  or  impossible  to  tell 
the  place  or  denomination  of  the  given  figure. 

Jllmtration.  The  3  of  the  number  3  24  may  mean  3  thousands,  or 
3  ten-thousands ;  but  when  the  space  between  the  3  and  2  is  filled,  the 
denomination  of  the  3  is  at  once  apparent.  Thus,  in  3024,  or  3124,  or 
3724,  the  3  represents  3  thousands;  but  in  30024,  31024,  32324,  or 
37824,  it  represents  3  ten-thousands. 

(b.)  The  digit  to  be  used  in  any  intervening  place  is  deter- 
mined by  the  number  of  units  to  be  represented  of  the  denom- 
ination of  that  place.  If  there  are  none,  then  zero  should  be 
used  ;  if  one,  then  1 ;  if  two,  then  2  ;  &c. 

Illustration.  In  order  that  9  may  represent  9  ten-millions,  it  must  be 
written  in  the  eighth  place,  at  the  left  of  the  point,  and  hence  there  must 
be  seven  figures  to  the  right  of  it.  If  we  wish  to  express  only  9  ten- 
millions,  or  90  millions,  the  intervening  places  must  be  tilled  by  zeroes, 
thus,  90,000,000  ;  but  if  we  wish  to  express  90  millions,  3  thousand,  8 
hundred,  and  7,  we  write  9  in  the  eighth  place,  as  before,  3  in  the  fourth, 
8  in  the  third,  7  in  the  first,  and  zeros  in  all  the  intermediate  places  i 
thus,  90,003,807. 

10.  Method  of  writing  Numbers. 
Examples. 
(a.)  In  writing  numbers  by  figures  we  may  begin  at  the 
left,  and  write  in  each  successive  period  the  figures  necessary  to 
express  the  required  number  of  units  of  the  denomination  of 
that  period ;  or  we  may  begin  at  the  right,  and  write  in  each 
successive  place  the  figures  expressing  the  required  number  of 
units  of  the  denomination  of  that  place,  taking  care  to  write  a 
zero  in  each  place  otherwise  unoccupied. 

(6.)  In  writing  numbers  be  careful  to  distinguish  tlie  decimal  point 
from  the  commas  used  to  separate  the  periods.  The  former  should  be  a 
dot,  so  carefully  made  that  it  cannot  be  mistaken  for  a  comma,  while  the 
latter  should  be  made  with  equal  care.  No  number  has  more  than  one 
decimal  point 


NOTATION   AND    NUMERATION  13 

(c.)  Accuracy  in  reading  and  writing  numbers  is  of  the  greatest  im- 
portance. If  the  numbers  used  in  solving  a  problem  are  copied  incor* 
rectly,  the  results  obtained  will  of  necessity  be  wrong ;  and  if  the  book 
from  which  the  problem  was  obtained  is  not  at  hand,  or  if  the  facts  and 
transactions  on  which  the  problem  was  based  are  forgotten,  it  will  be 
very  diflBcult,  and  usually  impossible,  tc  make  the  necessary  correc- 
tion. 

1.  Represent  by  figures  five  hundred  and  twenty-seven 
thousand,  four  hundred  and  eighty-nine. 

2.  Eight  thousand,  four  hundred  and  seven. 

3.  Eighty-five  thousand. 

4.  Eighty-five  thousand  and  one. 

5.  Eighty-five  thousand  and  thirty-one. 

6.  Nine  million,  eight  hundred  and  fifty-six  thousand,  seven 
hundred  and  twenty-one. 

7.  Twelve  million,  twelve  thousand,  and  twelve. 

8.  Four  billion,  eight  hundred  seventy-six  million,  five  hun- 
dred and  four  thousand,  three  hundred  and  one. 

9.  Four  billion,  eight  hundred  and  four  million,  eight  hun- 
dred and  four  thousand,  eight  hundred  and  four. 

10.  Thirty-seven  million,  eight  hundred  and  fifty-nine 
thousand. 

11.  Thirty-seven  billion,  eight  hundred  and  fifty-nine  mil- 
lion. 

12.  Thirty-seven  billion,  eight  hundred  and  fifty-nine 
thousand. 

13.  Thirty-seven  million,  eight  hundred  and  fifty-nine. 

14.  Forty  billion,  three  hundred  and  forty  million,  four  hun- 
dred and  eighty-seven  thousand,  five  hundred  and  nine. 

15.  Five  billion,  eight  hundred  and  seventy-six  thousand, 
seven  hundred  and  forty-six., 

16.  Seventy-five  trillion,  eight  hundred  and  seventy-six 
billion,  four  hundred  and  eighty-two  million,  four  hundred  and 
seventy-six  thousand,  three  hundred  and  twenty-seven. 

17.  Four  trillion,  seven  hundred  and  sixty-four  billion,  eight 
hundred  and  twenty-one  million,  six  hundred  and  seventeen 
thousand,  four  hundred  and  fifty-one. 

2 


H  NOTATION   AND   NUMERATION. 

18.  Seven  hundred  and  twenty-five  trillion,  eight  hundred 
and  seventy-six  billion,  four  hundred  and  three  million,  eight 
hundred  and  fifty  thousand,  four  hundred. 

19.  Three  hundred  and  six  trillion,  eighteen  billion,  four 
hundred  million,  three  thousand,  four  hundred  and  seventy-five. 

20.  Three  trillion,  three  hundred  and  ninety-nine  billion, 
three  hundred  and  ninety-nine  million,  three  hundred  thousand, 
four  hundred  and  three. 

2 1 .  Eighty-seven  trillion,  five  hundred  and  four  billion,  three 
hundred  million,  seven  thousand,  six  hundred  and  seventy-five. 

22.  Seventy  quadrillion,  seven  hundred  and  seven  billion, 
seven  thousand  and  seven. 

23.  Eighty-seven  quintillion,  eight  trillion,  seven  hundred 
billion,  eight  hundred  and  seventy  thousand,  and  eighty-seven. 

24.  Three  hundred  and  fifty-four  sextillion,  four  hundred 
and  seven  quintillion,  two  hundred  and  nine  quadrillion,  nine 
hundred  and  seventeen  trillion,  seven  hundred  billion,  eighty- 
six  million,  seven  thousand,  eight  hundred  and  fifty-two. 

25.  Seven  hundred  and  seven  quintillion,  two  hundred  and 
six  thousand. 

26.  Five  hundred  and  seventy  quadrillion,  five  hundred 
and  seventy. 

27.  Eight  sextillion,  eight  trillion,  eight  thousand,  and  eight. 

17.    Any   Gomhination  of  Figures  may  he  read  as  though 

alone. 

(a.)  Combinations  of  figures,  wherever  placed,  can  be  read  as 
though  they  stood  alone,  if  the  name  of  the  place  of  the  right 
hand  figure  be  given  after  reading  the  figures. 

For  instance,  347  always  stands  for,  and  may  be  read  as,  three 
hundred  and  forty-seven  of  the  denomtnation  of  the  place  occupied  by 
the  7. 

(6.)  To  illustrate  this  still  further,  we  have  written  opposite  each  of 
the  following  numbers  the  value  expressed  by  347  in  that  numb 

1.  347,241,         ...        347  thousands. 

2.  347,100,         .        .        .        347  thousands. 

3.  4,134,721,      .        .        .        347  hundreds. 

4.  23,476,258,675,      .        .        347  ten-millions. 


NOTATION    AND    NUMERATION. 


15 


(c.)  Let  the  pupil  give  the  value  expressed  by  the  409  in 
each  of  the  following  numbers,  and  also  the  value  expressed 
by  the  27  :  — 


1. 

40,927. 

6. 

568,340,927. 

2. 

274,090. 

7. 

40,927,534. 

3. 

8,172,764,096,134. 

8. 

27,409. 

4. 

52,706,409,273. 

9. 

4,092,768,375,674 

5. 

134,090,062,746. 

10. 

1,973,409,327,547 

18.    Analysis  of  Numbers. 

1.  What  is  the  greatest  number  of  tens  that  can  be  taken 
from  546,372  ? 

Ans.  54637  tens. 

2.  What  is  the  greatest  number  of  ten-thousands  that  can 
be  taken  from  53,075,423,697  ? 

Ans.   5307542  ten-thousands. 

What  is   the   greatest  number  of  hundreds  that  can  be 
taken 


3.  From  8,643,792  ? 

4.  From  27,948,673  ? 

5.  From  2,876? 

6.  Froij^  487,962? 


7.  From  79,762? 

8.  From  279,876,372  ? 

9.  From  25,986,137,953? 
10.  From  542,763? 


11.  What  is  the  greatest  number  of  thousands  that  can  be 
taken  from  each  of  the  above  numbers  ?  of  tens  ?  of  miUions  ?  * 
9f  billions  ?  of  hundred  thousands  ?  of  ten  millions  ? 

12.  Express  the  value  of  457869  in  hundreds  and  units. 
Am.  457869  equals  4578  hundreds  and  69  units. 

Express  the  value  of  each  of  the  following  in  hundreds  and 
units :  — 


13. 

8,796,784. 

17. 

6,972. 

14. 

86,724. 

18. 

79,843. 

15. 

7,807,375. 

19. 

987,509,875. 

16. 

46,739,725,876. 

20. 

570,072,307,679. 

*  If  any  number  is  less  than  a  million,  as  2876,  the  answer  may  be^ 
2876  i»  less  than  a  million^  and  therefore  no  millions  can  be  taken  from  it. 


16  NOTATION    AND    NUMERATIOlf. 

21.  Express  the  value  of  each  of  the  above  in  tens  and 
units.     In  ten-thousands  and  units. 

22.  Express  as  much  of  the  value  of  each  of  the  above  as 
you  can  in  ten-thousands  ;  as  much  of  the  remainder  as  you 
can  in  hundreds,  and  the  rest  in  units. 

Modd  of  Answer.  8796784  =  879  teu-thousands,  67  hundreds,  and  84 
units, 

23.  Express  as  much  of  the  value  of  each  of  the  above  as 
you  can  in  ten-millions,  as  much  of  the  remainder  as  you  can 


10,    Comparison  of  the  Values  represented  by  the  same  Figure 
in  different  Places, 

(a.)  Since,  as  we  have  seen,  the  figure  in  any  place  repre- 
sents ten  times  the  value  it  would  represent  if  written  one  place 
farther  towards  the  right,  one  hundred  times  the  value  it  would 
represent  if  written  two  places  farther  towards  the  right,  &c, 
it  follows  that  it  must  represent  one  *  tenth  of  the  value  it 
would  represent  if  written  one  place  farther  towards  the  left, 
one  one-*  hundredth  of  the  value  it  would  represent  if  written 
two  places  farther  towards  the  left,  &c. 

Note.  —  The  expression  "  Numbers  increase  from  right  to  left  in  a 
tenfold  ratio,"  is  not  a  true  statement  of  the  fact. 

A  unit  of  one  decimal  denomination  always  bears  the  same  ratio  to  a 
unit  of  the  next  higher  that  1  does  to  10 ;  but  the  ratio  which  the  value  of 
a  figure  of  one  decimal  denomination  bears  to  the  vahie  of  a  fiijure.  of 
the  next  higher  is  as  1  to  10  only  when  the  figures  are  alike.  For  in- 
stance, in  22  the  value  of  the  left  hand  figure  is  ten  times  that  of  the 
right  hand  figure,  while  in  25  it  is  only  four  times,  and  in  91  it  is  ninety 
times. 

Even  when  the  figures  are  alike  the  ratio  of  increase  is  not  tenfold. 
We  increase  a  number  by  adding  to  it.  To  increase  a  number  once,  one 
addition  must  be  made  to  it.  To  increase  it  twice,  two  additions  must 
be  made,  &c.    A  number  increased  by  once  itself  will  produce  twice  it- 

*  If  the  explanations  on  page  3d  are  not  suflRcient  to  enable  the  pupil 
to  understand  the  meaning  and  use  of  the  fractional  expressions  contained 
in  this  section,  he  should  study  the  first  part  of  the  section  on  fractions 
before  going  farther. 


AOrAriO!^    AND    NUMERATION.  17 

^If ,  increased  by  twice  itself,  will  produce  three  times  itself;  increased 
by  nine  times  itself,  will  produce  ten  times  itself;  increased  by  ten  times 
itself,  will  produce  eleven  times  itself,  &c.  Twenty  is  ten  times  two,  or 
ten  times  as  large  as  two,  but  is  only  nine  times  larger ;  for  if  we  make 
two  larger  by  nine  times  itself,  that  is,  if  we  add  nine  twos  to  two,  the 
result  will  be  twenty,  equal  to  ten  times  two. 

1.  Compare  the  values  expressed  by  the  4's  in  the  num- 
ber 444. 

Ans.  The  first  4  at  the  right  represents  one  tenth  of  the  value  of 
the  second  4,  and  one  one-hundredth  the  value  of  the  third  4 ;  the 
second  4  represents  ten  times  the  value  of  the  first  or  right  hand,  and  one 
tenth  of  the  value  of  the  third  or  left  hand  4  ;  the  third  4  expresses  ten 
times  the  value  of  the  second,  and  one  hundred  times  that  of  the  first. 

2.  Compare  the  values  expressed  by  the  6's  in  QQ ;  in  666 ; 
in  606  ;  in  660. 

3.  Compare  the  values  expressed  by  the  8's  in  88  ;  in  880 ; 
in  808  ;  in  8888  ;  in  8008. 

4.  Compare  the  values  expressed  by  the  3's  in  333 ;  in 
3030  ;  in  3303  ;  in  333333. 

20.    Places  at  Right  of  Decimal  Point. 

(a.)  In  conformity  with  the  same  principle,  a  figure  written 
in  the  first  place  at  the  right  of  the  point  would  represent  tenths 
of  units,  or  simply  tenths ;  one  written  in  the  second  place  at 
the  right  would  represent  hundredths  of  units,  or  simply  hun- 
dredths, &c. 

(b.)  Hence,  the  places  at  the  right  of  the  point  have  been 
name'd  as  indicated  below. 

i  ^ 

=  1    .1 

.000000000 

£    d    a  S  S  ~  S  S  S 

a, '^ '^  o.  o.  P.  Q.  a.  a, 


18 


NOTATION    AND    NUMERATION. 


1.   Which  place  from  the  point  is  occupied  by  the  milliontha 
figure  "i 

Ans.  The  sixth  place  at  the  right 

Which  place  from  the  point  is  occupied  by  the 


2. 

Tenths'  figure  ? 

6. 

Thousandths'  ? 

8. 

Hundi-edths'? 

7. 

Billionths*? 

4. 

Ten-thousandths'  ? 

8. 

Hundred-thousandths'  f 

5. 

MiUionths'? 

9. 

Hundred-millionths'  ? 

10.   What  would  be  the  denomination  of  a  figure  in  the 
second  place  at  the  right  of  the  point  ? 


11.  In  the  fifth? 

12.  In  the  sixth? 

13.  In  the  fourth? 

14.  In  the  seventh? 


15.  In  the  first? 

16.  In  the  ninth? 

17.  In  the  eighth  ? 

18.  In  the  third? 


91  •     Method  of  reading  Numbers  expressed  hy  Figures  at 
Right  of  Decimal  Point. 

(a.)  Numbers  expressed  by  figures  written  at  the  right  of 
the  point  are  read  on  exactly  the  same  principles  as  those 
expressed  by  figures  at  the  left. 

Thus,  if  we  wish  to  read  .37,  we  observe  that  the  right  hand  figure  is 
in  the  hundredths'  place.  We  read  it  then  as  though  written  thirty- 
seven  one  hundredths,  or  -^^jj.  The  following  will  furnish  further  il- 
lustrations of  the  principle  involved :  — 


.86      =.^%%. 
.0086  =  T-AW- 

.0349  =  ^nu- 

3.49  rr|*9  =3t^V 
6.07  =m  =6^^^. 
10.1    =-W    =10^^. 

7.  30.03  =  2i^^o/ =  SO^g^. 

8.  7.008706  1=  UUm  =  7t^Vt/»^W 
(6.)    Read  the  following,  expressing  the  value  in  all 

where  it  is  greater  than  unity,  both  as  an  improper  fraction 
and  as  a  mixed  number :  — 


9. 

.086  =  T-«e-ii. 

10. 

.349  =.  t^oV^. 

11. 

.000349  =  ^^g4§^^. 

12. 

84.9=W  =  34A. 

13. 

l.Ol^lU^Mzr. 

14. 

300.3  =  ^%^  =  300^5^. 

15. 

3.003  =  |8H=3^^^^. 

NOTATION    AND    NUMERATION. 


19 


16. 

.73 

26. 

50.05 

17. 

.08 

27. 

0.0764 

18. 

.798 

28. 

37.9427 

19. 

6.4 

29. 

876.5874 

20. 

7.0037 

30. 

3299.4856328 

21. 

4.287 

31. 

1000000.0000001 

22. 

.940094 

32. 

87.203794 

23. 

3.06006 

33. 

87.000000079 

24. 

40.7000407 

34. 

.000000007 

25. 

21.3304206 

35. 

2006.000002006 

2S.    Decimal  Fractions,  Method  of  writing  them. 

(a.)  Any  fraction  whose  denominator  is  a  power*  often  may 
be  expressed  by  writing  the  numerator  so  that  its  right  hand 
figure  shall  occupy  the  place  of  the  same  name  with  its  de- 
nominator, and  omitting  the  denominator.  Fractions  thus 
written  are  called  Decimal  Fractions,  to  distinguish  them  from 
Vulgar  Fractions,  or  those  whose  numerator  and  denominator 
are  both  written. 

Note.  —  The  first  fifteen  examples  of  21  are  written  both  as  decimal 
and  vulgar  fractions,  while  the  last  twenty  are  only  written  as  decimal 
fractions. 

The  greatest  liability  to  error  in  writing  decimal  fractions  is  in 
placing  the  point.  The  learner  should  bear  this  in  mind,  and  take  pains 
to  guard  against  it,  remembering  that  if  the  point  is  not  correctly 
placed,  each  figure  of  the  number  will  express  a  wrong  value. 

{b.)    Write  the  following  in  the  form  of  decimal  fractions :  — 

1.  Eight  tenths. 

2.  Fifty-four  hundredths. 

3.  Eight  hundred  and  seventy-five  thousandths. 

4.  Fifty-four  thousandths. 

5.  Eight  hundred  and  sixty-four  ten-thousandths. 

6.  Six,  and  eighty-seven  hundredths. 

7.  Six,  and  eighty-seven  ten-thousandths. 

8.  Four  hundred  and  thirty-seven  tenths. 


*  For  definition  of  the  word  power,  see  article  105,  (d.) 


20 


NOTATION    AND    NUMERATION. 


9.  Three  hundred  and  four  thousands,  and  three  hundred 
and  four  tliousandths. 

10.  Three  hundred  and  four  thousands,  and  three  hundred 
and  four  millionths. 


(c.)   Write  th( 

3fol 

owin^ 

;  in  the  form  of  decimal  fractions  :  — - 

11.     /^. 

21. 

l^D- 

31.     i^/ZcAr. 

12.    ^v 

22. 

T^ffXX- 

32.     T§8iir- 

13.     T§^. 

23. 

^J- 

33.     rxj^inj- 

14.     i^AV 

24. 

m- 

34.     ruf^Tju- 

15.     tUtj- 

25. 

TuVuir- 

35.     fn§^§. 

16.     HU- 

26. 

!§i*l- 

36.     TTiwSjJSjy 

17.     3^^. 

27. 

97y4^. 

37.     500^5^. 

18.     40^*^. 

28. 

SOJ^. 

38.     50^^^^. 

19.     24^V 

29. 

gyf^- 

39.     5j^^^^. 

20.     4^«,. 

30. 

17t'uV 

40.     637t^6^V^^. 

41.     46948f^V(j'^ 

48. 

185476-,^VitW 

42.  mum^- 

49. 

342y^^3^^/_^;j^. 

43.     8270000^^^^VuTrc5 

. 

50. 

^OOQxjjmh^jj' 

44.     ^Jf^Vo^-^. 

51. 

-^w^ll^. 

45.     IxTrirWTr- 

52. 

1000000t^,,^^o-. 

46.     6000^^-^. 

53. 

487t^o^VTTV^. 

47.  7200000000( 

^lOooljjmTUJj- 

54. 

851tVit^^^. 

33.     Exercises  in  determining  Position  of  Point. 

1.  Where  must  the  point  be  placed  in  order  that  the  figures 
746  may  represent  tenths? 

Ans.   Between  the  4  and  6;  thus,  74.6 

2.  Where  must  the  point  be  placed  in  order  that  the  figures 
746  may  represent  millionths  ? 

Ans.  Six  places  to  the  left  of  the  6 ;  wliich  requires  that  three  places 
be  filled  with  zeros ;  thus,  .000746 

3.  Where  must  the  point  be  placed  in  order  that  the  figures 
573  may  express  573  tenths?  hundredths?  units?  tens?  hun- 
dreds? thousands?  thousandth.^?  millions?  millionths? 


NOTATION    AND    NUMERATION.  21 

4.  "Where  must  the  point  be  placed  in  order  that  the  figures 
87064  may  express  87064  hundreds  ?  hundredths  ?  ten-thou- 
sands ?  ten-thousandths  ?  hundred-millions  ?  hundred-mil- 
lionths  ?   tens  ?   tenths  ? 

5.  Where  must  the  point  be  placed  in  order  that  the  figures 
497837  may  express  497837  units?  millions?  billionths? 
hundredths?   millionths?   hundreds?   hundred-thousandths? 

^4.     Effect  of  changing  Place  of  Point. 
Multiplication  and  Division  by  Powers  of  10. 

{a.)  Since  (8,  19)  each  figure  of  a  number  represents  10 
times  the  value  which  it  would  represent  if  written  one  place  far- 
ther towards  the  right,  100  times  the  value  it  would  represent  if 
written  two  places  farther  towards  the  right,  &c.,  and  -^  of  t&e 
value  it  would  represent  if  written  one  place  farther  towards 
the  left,  Yuu  of  the  value  it  would  represent  if  written  two 
places  farther,  &c.,  it  follows  that  removing  the  figures  repre- 
senting a  number  one  place  farther  towards  the  left,  or,  which 
is  the  same  thing,  removing  the  point  one  place  to  the  right, 
multiplies  the  number  by  10,  while  removing  the  figures  one 
place  to  the  right,  or  the  point  one  place  to  the  left,  divides  it 
by  10.  A  change  of  two  places  would  in  like  manner  multi- 
ply or  divide  by  100,  of  three  places  by  1000,  &c. 

These  principles  generalized  would  be  stated  thus :  — 

(6.)  To  express  the  product  of  any  numher  multiplied  hy  any 
power  q/"  10,  remove  the  decimal  point  as  many  places  to  the 
right  as  there  are  zeros  used  in  writing  the  given  midtiplier. 

(c.)  To  express  the  quotient  of  a  number  divided  by  any 
power  of  \0,  remove  the  decimal  point  as  many  places  to  the 
left  as  there  are  zeros  used  in  writing  the  given  divisor. 

{d.)  When,  by  such  cliange,  any  places  between  the  numher 
and  point  are  left  vacant,  they  must  be  filled  by  zeros. 

Note.  —  The  rule  usually  given  for  multiplying  by  the  powers  of  10, 
is  to  "  annex  as  many  zeros  to  the  multiplicand  as  there  are  zeros  in  the 
multiplier."  It  is,  however,  a  very  defective  one,  as  it  only  applies  when 
the  decimal  point  is  not  written,  and  then  only  multiplies  the  number  by 


23 


NOTATION    AND    NUMERATION. 


changing  the  place  to  which  we  refer  the  point.  It  is  the  more  cbjeo 
tionable,  as  it  tends  to  convey  the  false  idea  that  the  zero  is  essential  to 
the  multiplication.  As  a  matter  of  fact,  annexing  any  figure  whatever 
to  a  number  will,  if  the  decimal  point  is  omitted,  multiply  it  by  10,  for 
it  will  change  the  place  to  which  the  decimal  point  is  referred ;  but  if 
the  annexed  figure  is  other  than  zero,  its  value  will  be  added  to  the  prod- 
uct of  the  number  by  10.  Thus,  annexing  7  to  43  gives  437,  which 
equals  10  times  43,  plus  7. 

The  principle  involved  is  this :  that  every  change  which  is  made  in  the 
position  of  figures  with  reference  to  the  decimal  point,  —  whether  it  is 
made  by  changing  the  position  of  the  point,  or  by  writing  other  figures 
between  the  given  figures  and  the  point,  —  alters  the  value  they  repre- 
Bent,  by  multiplying  or  dividing  them  by  10  or  some  power  of  ten.  A 
figure  can  only  alter  the  value  expressed  by  other  figures  when  it  is 
written  between  them  and  the  point. 

(e.)  How  will  you  express  in  figures  the  results  of  the  fol- 
lowing indicated  operations  ? 

14.  4279  X  1000 

15.  .6307  X  100 

16.  .00694  X  10 

17.  5429  H-  100 

18.  400.794  -^  10 

19.  .0054279  —  10000 

20.  .0004  X  1000000 

21.  .002  X  10000 

22.  348.796  X  100 

23.  2  X  1000000 

24.  2  -^  1000000 

25.  .006  X  1000000000 

26.  .006  -i-  1000000000 

(/.)  Let  the  student  now  tell  by  inspection,  without  chan* 
ging  the  place  of  the  point  or  re-writing  the  numbers,  the  re- 
sult of  the  above  indicated  operations. 


1. 

87  X  10* 

2. 

.87  X  10 

3. 

.0087  X  10 

4. 

87-T-lO 

5. 

.087  -r- 10 

6. 

870  -^ 10 

7. 

5.7  -f- 100 

8. 

.057  -^  100 

9. 

.3278  X  100 

10. 

4.5786  X  1000 

11. 

.4-^10 

12. 

479.643  X  1000 

13. 

479.643  -r- 1000 

♦  The  student  should  remember  that  when  the  decimal  point  is  not 
marked,  it  is  always  understood  to  belong  at  the  right  of  the  given  figures. 


NOTATION    AND    NUMERATION.  23 

Thus  .  eighty-seven  multiplied  by  ten  equals  eight  hundred  and  severity  ; 
eighty-seven  hundredths  multiplied  by  ten  equals  eighty-seven  tenths,  or  eight 
and  seven  tenths,  &c.  He  should  learn  to  do  this  without  the  slightest 
hesitation. 

^5      Change  of  Denomination. 

(a.)  The  value  of  a  number  may  be  expressed  in  terms  of 
any  other  decimal  denomination  as  well  as  in  units,  bj  making 
the  requisite  change  in  the  place  of  the  point.     Thus  :  — 

847  =  34.7  tens,  =  8.47  hundreds,  =  .847  of  a  thousand,  =  .0847  of 
a  ten-thousand,  &c. 

847  =  8470  tenths,  =  84700  hundredths,  =  847000  thousandths,  &c 
642.06  =  6.4206  hundreds,  =  64206  hundredths,  &c 

(b.)  Express  the  value  of  each  of  the  following  in  tenths, 
then  in  tens ;  in  hundredths,  then  in  hundreds ;  in  millionths, 
and  then  in  millions :  — 


1. 

4327 

4. 

2700 

7. 

4683.7G42 

2. 

82794.6 

5. 

.048 

8. 

.00006 

3. 

.437 

6. 

2.7 

9. 

8000000 

2G.    French  Method  of  Numeration. 

The  foregoing  method  of  numeration  is  called  the  French 
method.  It  divides  the  figures  expressing  a  number  into 
periods  of  three  figures  each,  making  a  unit  of  one  period 
equal  to  one  thousand  units  of  the  next  lower  period. 

Thus  one  million  equals  one  thousand  thousands  ;  one  billion  equals 
one  thousand  millions  ;  one  trillion  equals  one  thousand  billions,  &c. 

27.    English  Method  of  Numeration. 

(a.)  There  is  another  method,  called  the  English  method, 
which  divides  the  figures  expressing  a  number  into  periods  of 
six  figures  each,  making  a  unit  of  one  period  equal  to  one 
miUion  units  of  the  next  lower  period. 

(h.)  By  this  method  one  billion  equals  one  million  millions ; 
one  trillion  equals  one  million  billions,  &c.  This  is  illustrated 
in  the  following  example :  — 


2Jl  notation  and  numeration. 


9,7  5  27  30,665  897,2530  60,93457  8 


il  °J  *J  J 

IP  n  a  fl 

H  «  a  p 

(c.)  In  this  method,  as  in  the  French,  we  read  the  figures 
in  each  period  as  though  they  stood  alone,  caUing  afterwards 
the  name  of  the  period. 

{d.)  The  above  number  would  be  read,  9  quadrillions,  752730  tril- 
lions, 665897  billions,  253060  millions,  934578. 

38.     Comparison  of  French  and  English  Methods. 

(a.)  It  will  be  readily  seen  that  while  the  English  periods 
bear  the  same  name  as  the  French,  and  while  one  thousand 
and  one  million  represent  the  same  number  in  the  two  sys- 
tems, one  billion,  one  trillion,  or  a  unit  of  any  higher  denomina- 
tion is  much  greater  in  the  English  system  than  in  the  French. 

Thus,  an  English  billion  equals  a  French  trillion ;  an  English  trillion 
equals  a  French  quintillion. 

(h.)  The  French  method  is  the  one  generally  used  in  this  country  and 
on  the  continent  of  Europe,  and  being  much  more  convenient  than  the 
English,  has  been  adopted  in  part  in  England,  and  is  likely  to  come  into 
general  use  there. 

39.    Numbers  to  be  read  according  to  the  English  Method, 

1.  426,794798,764387. 

2.  86432,795876,942759. 

3.  237000,568975,006723. 

4.  6,456327,309670,800659. 

6.     325,897563,475003,90OO65.65400a. 


NOTATION    AND    NUMERATION.  25 

30.    The  Roman  Method. 

(a.)  The  Roman  method  of  notation  represents  numbers  by 
letters  of  the  alphabet. 

(6.)  It  is  now  chiefly  used  in  numbering  the  sections  or  chapters  of  a 
book,  the  pages  of  a  preface  or  introduction,  the  year  of  the  Christian 
era,  or  when  it  is  necessary  to  distinguish  one  class  of  numbers  from 
another. 

(c.)   The  letters  used  are  the  following,  viz. :  — 

I ,  which  stands  for  One. 

V ,  which  stands  for  Five. 

X ,  which  stands  for  Ten. 

L ,  which  stands  for  Fifty. 

C ,  which  stands  for  One  Hundred. 

D ,  which  stands  for  Five  Hundred. 
M ,  which  stands  for  One  Thousand. 

(rf.)  Other  numbers  are  represented  by  repetitions  and 
combinations  of  these  letters. 

(e.)  When  a  letter  is  repeated,  it  indicates  that  the  number 
it  represents  is  to  be  repeated. 

Thus:  II.  =  two  ;  HI.  =  three  ;  XX.  =  twenty  ;  XXX.  =  thirty,  &c. 

(/.)  If  a  letter  expressing  one  number  be  placed  before  a 
letter  expressing  a  larger  number,  the  former  is  to  be  sub- 
tracted fix)m  the  latter;  but  if  the  letter  expressing  the  larger 
value  be  placed  first,  the  values  of  the  two  are  to  be  added 
together. 

Thus  .  IV.  =  four ;  IX.  =  nine  ;  XL.  =  forty,  &c. ;  while  VI.  =  six ; 
XI.  =  eleven ;  LX.  =  sixty,  &c. 

(g,)    In  the  following  columns,  the  letters  stand  for  the 
numbers  written  against  them :  — 
I.  .  One. 
II.  .  Two. 

III.  .  Three. 

IV.  .  Four. 
V.  .  Five. 

VI.  .  Six. 


VII. 

.  Seven. 

V^III. 

.  Eight. 

IX. 

.  Nine. 

X. 

.  Ten. 

XL 

.  Eleven. 

XII. 

.  Twelve. 

d« 


NOTATION    AND    NUMERATION. 


XIII. 

.  Thirteen. 

L. 

.  Fifty. 

XIV. 

.  Fourteen. 

LX. 

.  Sixty. 

XV. 

.  Fifteen. 

LXX. 

.  Seventy. 

XVI. 

.  Sixteen. 

LXXX. 

.  Eighty. 

XVII. 

.  Seventeen. 

xc. 

.  Ninety. 

XVIII. 

.  Eighteen. 

XCIX. 

.  Ninety-nine. 

XIX. 

.  Nineteen. 

c. 

.  One  Hundred. 

XX. 

.  Twenty. 

CL. 

.  One   Hundred   and 

XXL 

.  Twenty-one. 

Fifty. 

XXII. 

.  Twenty-two. 

CLXXXVin. .  One  Hundred 

XXIII. 

.  Twenty-three. 

and  Eighty-eight 

XXIV. 

.  Twenty-four. 

CC. 

.  Two  Hundred. 

XXV. 

.  Twenty-five. 

CCC. 

.  Three  Hundred. 

XXVI. 

.  Twenty-six. 

CD. 

.  Four  Hundred. 

XXVII. 

.  Twenty-seven. 

D. 

.  Five  Hundred. 

srxviii. 

.  Twenty-eight. 

DC. 

.  Six  Hundred. 

XXIX. 

.  Twenty-nine. 

DCC. 

.  Seven  Hundred. 

XXX. 

.  Thirty. 

DCCC. 

.  Eight  Hundred. 

XXXI. 

.  Thirty-one. 

CM. 

.  Nine  Hundred. 

XL. 

.  Forty. 

M. 

.  One  Thousand. 

XLL 

.  Forty-one. 

MDCCCLIV.  .  1854. 

XLIX. 

,  Forty -nine. 

(h.)  A  dash  placed  over  a  letter  makes  it  express  thousands 
instead  of  ones.  Thus,  Y.  =  5000;YI.=  6000  ;X.  =  50,000; 
XC.=  90,000,  &c. 

(t.)    Read  the  following  numbers  :  — 

XCVIII. 
DCCXLII. 
MDCCLXXVI. 
XXDCCCXCIX. 


CCXLVIIICCCL. 


DCLIDLXXL 
DCCCXCIXCCCXXXHL 


TABLES    OF   MONEY,    WEIGHTS,    AND    MEASURES.         27 

SECTION   III. 
TABLES  OF  MONEY,  WEIGHTS,  AND  MEASURES. 

31.    Introductory. 

{a.)  All  nations,  excepting  perhaps  the  most  barbarous,  have 
some  kind  of  money ;  but  each  nation  has  a  system  peculiar 
to  itself,  and  generally  differing  from  every  other  in  its  de- 
nominations, coins,  &c.  The  people  of  the  United  States 
reckon  money  in  dollars  and  cents,  the  English  reckon  it  in 
pounds,  shillings,  and  pence,  and  the  French  in  francs  and 
centimes. 

(h.)  So  too  each  nation  has  a  peculiar  system  of  weights  and 
measures,  some  of  the  most  important  of  which  are  illustrated 
in  the  following  tables. 

3S.     United  States,  or  Federal  Money. 
(a.)   The  money  of  the  United  States  is  called   Federal 
Money. 

TABLE    OF   FEDERAL    MONEY. 

10  mills      r=  1  cent. 

10  cents     =  1  dime. 

10  dimes    •=.  1  dollar. 

10  dollars  =  1  eagle. 

(5.)  The  coins  of  the  United  States  are  the  dollar,  the  half 
dollar  or  fifty-cent  piece,  the  quarter  dollar  or  twenty-five-cent 
piece,  the  dime  or  ten-cent  piece,  the  half  dime  or  five-cent 
piece,  the  three-cent  piece,  the  cent ;  the  eagle  or  ten  dollar 
piece,  the  double  eagle  or  twenty-dollar  piece,  the  half  eagle 
or  five-dollar  piece,  and  the  quarter  eagle,  worth  two  and  a 
half  dollars. 

(c.)  The  dollar  is  coined  both  of  gold  and  silver ;  the  coins 
worth  more  than  a  dollar  are  of  gold,  and  the  others,  with  the 
exception  of  the  cent,  are  of  silver.     The  cent  is  of  copper. 


28  TABLES    OF    MONKY,    WEIGHTS,    AND    MEASURES. 


(d.)  Values  in  federal  money  are  usually  expressed  in  dol- 
lars and  cents,  or  in  dollai-s,  cents,  and  mills,  the  dollar  being 
regarded  as  the  unit,  the  cent  as  one  hundredth  of  it,  and  the 
mill  as  one  tenth  of  the  cent,  or  one  thousandth  of  the  dollar. 
Indeed,  dimes,  cents,  and  mills  being  nothing  more  than  tenths^ 
hundredths,  and  thousandths  of  a  dollar,  the  figures  represent- 
ing them  may  be,  and  usually  are,  written  in  the  form  of  a 
decimal  fraction,  and  may  with  perfect  propriety  be  read  as 
such. 

(e.)  The  character  $,  placed  at  the  left  of  figures,  shows 
that  they  stand  for  dollars. 

Illustrations,  (a.)  $3.75  =  *3  dollars  and  75  cents,  =  3  dollars  and 
J7  -^  of  a  dollar,  =  3  dollars,  7  dimes,  and  5  cents,  &c. 

(6.)  $237,264  =  *  237  dollars,  26  cents,  and  4  mills,  =  237  dollars 
and  -2q?^7^  of  a  dollar,  =  ^.^.^H*  of  a  dollar,  =  23726.4  cents,  = 
2372G4  mills,  &c. 

(c.)  $7,042  =  *  7  dollars,  4  cents,  and  2  mills,  =  7  dollars,  42  mills, 
=  7  dollars  and^^of  a  dollar,  =^^4|  of  a  dollar,  =  7042  mills,  ==• 
704.2  cents,  &c- 

Exercises. 

(/,)  Read  each  of  the  following,  giving  the  value  1st  in  dol- 
lars and  cents,  or,  where  there  are  mills,  in  dollars,  cents,  and 
mills ;  2d,  in  dollai^s  and  decimal  parts  cf  a  dollar ;  3d,  in 
cents ;   4th,  in  mills :  — 

$  .073  13.  $  7084.79 

$  30.07  14.  $  400.06 

$  2587.00  lo.  $  .125 

$  25.00  16.  $  97.886 

$  19.875  17.  $  20.07 

$  .625  18.  $  .06 

English  Money, 


1.     $  6.79 

7. 

2.     $  8.03 

8. 

3.     $  764.37 

9. 

4.     %  28.00 

10. 

5.     $  5.976 

11. 

6.     $  .073 

12. 

33. 

{a.)    The  mone; 

irused 

ling  Money. 

*  The  form  marked  with  a  star  represont'S  the  usual  method  of  reading 
these  values. 


TABLES    OF    MONEY,    WEIGHTS,    AND    MEASURES.         29 
TABLE    OF    STERLING   MONEY. 

4  farthings  =  1  penny. 
12  pence       =  1  shilling. 
20  shillings  =  1  pound. 
(b.)    When  numbers  expressing  values  in  sterling  money 
are  used,  the  character  £  marks  the  pounds,  s.  the  shillings, 
d.  the  pence,  and  qr.  the  farthings. 

Illustration.    27  pounds,  18  shillings,  5  pence,  and  3  farthings,  may 

£       B.    d.  qrs. 

be  written  thus  :  28£,  18  s.  5  d.  3  qrs. ;  or  thus :  27  18  5  3.  Farthings 
are  also  frequently  expressed  as  fourths  of  a  penny.  Thus,  7  d.  3  qrs.  =« 
7|d.,  and  both  forms  may  be  read  as  7  pence  and  3  farthings. 

(c.)  The  coin  representing  the  pound  is  called  the  sover- 
eign. Its  value  in  United  States  money  varies  from  four 
dollars  and  eighty-three  cents  to  four  dollars  and  eighty-six 
cents,  but  is  usually  about  four  dollars  and  eighty-four  cents. 
There  are  several  other  coins  used  in  England,  of  which  we 
will  mention  only  two,  viz.,  the  guinea,  or  twenty-one  shilling 
piece,  and  the  crown,  or  five-shilling  piece. 

34  •    Avoirdupois    Weight. 

(a.)  Almost  all  articles,  except  gold,  silver,  and  jewels,  are 
weighed  by  what  is  called  Avoirdupois  Weight. 

TABLE    OF    AVOIRDUPOIS   WEIGHT. 

16  drams  =  1  ounce. 

16  ounces  =  1  pound. 

25  pounds  =  1  quarter. 

4  quarters  =  1  hundred  weight. 

20  hundred  weight  =  1  ton. 

(b.)  The  abbreviations  made  use  of  in  this  weight  are 
T.  for  ton,  cwt.  for  hundred  weight,  qr.  for  quarter,  lb.  for 
pound,  oz.  for  ounce,  and  dr.  for  dram. 

T.  cwt  qr.    lb.  oz.  dr. 
Forexample,  5T.l7cwt.3qr.  13lb.  8oz.  7dr.  =  5    17    3    13    8    7  = 

5  tons,  17  hundred  weight,  3  quarters,  13  pounds,  8  ounces,  and  7  drams. 

(c.)   Formerly  the  quartei  was  reckoned   at  28  pounds,  the  hundred 

a* 


80         TABLES    OF    MONET,    WEIGHTS,    AND    MEASURES. 

weight  at  112  pounds,  and  the  ton  at  2240  pounds,  and  they  arc  so 
reckoned  at  the  present  time  in  Great  Britain  as  well  as  in  the  standard 
of  the  United  States  government.  Most  of  the  states  of  the  Union 
have,  however,  passed  laws  fixing  the  values  as  in  the  table,  and  they 
are  almost  always  so  reckoned  by  merchants  in  buying  and  selling. 

35,    Troy   Weight. 

(a.)  The  weight  used  in  weighing  gold,  silver,  and  precious 
stones  is  called  Troy  Weight.  This  weight  is  also  used  in 
philosophical  experiments. 

TABLE    OP    TROT    WEIGHT. 

24  grains  =z  1  pennyweight. 

20  pennyweights  =  1  ounce. 
12  ounces  =  1  pound. 

(b.)    In  this  weight,  lb.  stands  for  pound,  oz.  for  ounce, 

dwt.  for  pennyweight,  and  gr.  for  grain. 

Illustration.     13  pounds,  7  ounces,   18  pennyweights,  23  grains,  may 
"*  lb.   oa.  dwt    gr. 

be  expressed  thus  :  13  lb.  7  oz.  18 dwt.  23 gr.;  or  thus:  13   7    18   23. 

(c.)  The  "  carat,"  which  equals  four  grains,  is  used  in  weighing  dia- 
monds. The  term  cai'at  is  also  used  in  stating  the  fineness  of  gold,  and 
means  the  twenty-fourth  part  of  any  weight  of  gold  or  gold  alloy.  Pure 
gold  is  "  24  carats  fine."  Gold  is  22  carats  fine  when  ^2  of  it  is  pure 
gold  and  ^^  is  alloy. 

30 .    Apothecaries'    Weight, 
(a.)    Tlie  weight  used  in  compounding  or  mixing  medicines  is 
called  Apothecaries'  Weight.     Physicians  write  their  prescrip- 
tions in  this  weight,  but  medicines  are  bought  and  sold  by 
Avoirdupois  Weight. 

TABLE    OF   APOTHECARIES*    WEIGHT. 

20  grains     r=:  1  scruple. 
3  scruples  =  1  dram. 
8  drams     =  1  ounce. 
12  ounces     =  1  pound. 
To  mark  the  denominations  of  this  weight,  we  use  the  fol- 
lowing cliaracters,  viz. :  It)  for  pound,  S  for  ounce,  3  for  dram, 
9  for  scruple,  and  gr.  for  grain. 


TABLES    OB^    MONET,    WjEIGIITS,    AND    MEASURES.         3l 

For  example:  5  fc  6  5  4  3  29  17grs.,  would  be  read  as  5  pounds, 
6  ounces,  4  drams,  2  scruples,  and  17  grains. 

S7,    Compa?'ison  of  Troy,  Avoirdupois,  and  Apothecaries* 
Weights. 

{^a.)  The  only  difference  between  Troy  Weight  and  Apoth- 
ecaries' Weight  is,  that  in  the  former  the  ounce  is  divided  into 
pennyweights  and  grains,  while  in  the  latter  it  is  divided  into 
drams,  scruples,  and  grains.  The  pound,  ounce,  and  grain  are 
the  same  in  both  weights. 

{b.)  The  value  of  denominations  of  the  same  name  in 
Avoirdupois'  and  Troy  Weights  differs  very  materially,  as 
may  be  seen  from  the  following  table,  which  shows  the  value 
in  Troy  grains  of  each  denomination  we  have  given  in  the 
preceding  tables  of  weights. 


TABLE 

:  OF 

COMPARISON 

', 

1  lb.  Av. 

=  7000 

gr. 

Troy. 

1  lb.  Tr.  = 

lib 

=  5760 

(( 

u 

1  oz.  Av. 

==  437i 

(( 

(t 

1  oz.  Tr.  = 

li 

=  480 

u 

u 

1  dr.Av. 

=  27^* 

ii 

ii 

1   5 

—  60 

a 

it 

1  9 

=  20 

a 

it 

1  dwt. 

=  24 

a 

it 

1  gr.  Ap. 

=  1 

a 

u 

(c.)    From  the  above  table  we  should  find  by  calculation 
that 

144  lb.  Av.  =  175  lb.  Tr., 
and  192  oz.  Av.  =  175  oz.  Tr. 
Therefore,  1  lb.  Av.  =  |J|  or  1^\\  lb.  T\, 
and  1  oz.  Av.  z=  -}^^|  oz.  Tr. 

(d,)  Which  is  the  heavier,  and  why, — 

1.  A  pound  of  gold  or  a  pound  of  feathcri*s.^ 

2.  A  pound  of  lead  or  a  pound  of  feathers  ? 

3.  A  pouud  of  gold  or  a  pound  of  lead  ? 


32  TABLES    OF    MONEY,    WEIGHTS,    AND    MEASURES. 

4.  An  ounce  of  gold  or  an  ounce  of  feathers  ? 

5.  An  ounce  of  lead  or  an  ounce  of  feathers  ? 

6.  An  ounce  of  lead  or  an  ounce  of  gold  ? 

38.    Long  Measure. 

(a.)    Distances  in   any  direction  are  measured  by  Long 
Measure. 

TABLE    OF    LONG   MEASURE 


12 

lines 

— 

1  inch. 

12 

inches 

rz: 

1  foot. 

3 

feet 

■z=. 

1  yard. 

i)^  yards,  oi 
16^  feet 

— 

1  rod  or 

pole 

40 

rods 

z=z 

1  furlong. 

8 

furlongs 

= 

1  mile. 

3 

miles 

z= 

1  league. 

(J.)  In  this  measure,  le.  stands  for  league,  m.  for  mile,  fur. 
for  furlong,  rd.  for  rod,  yd.  for  yard,  ft.  for  foot,  and  in.  foi 
inches. 

(c.)  Surveyors  usually  measure  distances  by  means  of  a 
chain  4  rods  in  length,  called  Gunter's  chain,  i  the  Survey- 
or's chain.  This  chain  contains  100  equal  links ;  25  links 
will,  therefore,  equal  1  rod,  and  1  link  will  equal  7f  f  inches. 

39.    Cloth  Measure, 
(a.)   This  measure  is  used  for  measuring  cloth,  ribbons,  &c 

TABLE    OF    CLOTH   MEASURE. 

2^  inches      =  1  nail. 
4     nails        =  1  quarter. 
4     quarters  =  1  yard. 

(3.)  In  this  measure,  yd.  stands  for  yard,  qr.  for  quarter, 
and  na.  for  nail. 

(c.)  The  yard  and  inch  are  the  same  in  length  as  the  yard 
and  in':i^  r  Long  Measure 


TABi.KS    OF    MONEY,    WKIGHTS,    AND    MEASURES. 


3t 


40.    Square  Measure. 

(a.)    Square  Measure  is  used  for  measuring  surfaces. 

As  true  ideas  of  the  nature  of  the  right  angle,  rectangle,  and  square 
are  essential  to  a  just  appreciation  of  this  measure,  we  insert  the  follow- 
ing definitions  and  illustrations,  leaving  it  with  the  teacher  to  give  such 
others  as  he  may  deem  necessary. 

(b.)  When  two  lines  meet  at  a  point,  their  difference  in 
direction  is  called  the  angle  of  the  two  lines.  The  point 
'•here  they  meet  is  called  the  vertex  of  the  angle. 

Fig.  1. 

(c.)  In  this  figure,  the  lines  marked  B  A 
and  B  C  form  an  angle  whos.6  vertex  is  at  B. 

In  reading  an  angle,  the  letter  at  the  vertex, 
is  always  made  the  middle  one.  The  angle  in 
Fig.  1  may  he  read  either  as  the  angle  ABC, 
or  as  the  angle  C  B  A. 


(d.)  In  this  figure  there  are  two  an- 
gles, viz.,  D  B  A  and  A  B  C.  The 
first  is  the  difference  in  direction  of  the 
two  lines  A  B  and  D  B,  and  the  second 
is  the  difference  in  direction  of  the  two 
lines  A  B  and  B  C.  It  is  evident  that 
the  first  angle  is  larger  than  the  second. 

(e.)  When  the  two  angles  formed  by  one  straight  line 
meeting  another  are  equal  to  each  other,  they  are  called  right 
angles^  and  the  two  lines  are  said  to  be  perpendicular  to  each 
other 


Fig.  3. 
D 


(/)  Suppose  that  the  straight  line  D  B 
should  so  meet  the  straight  line  A  C  as  to  make 
the  adjacent  angles  A  B  D  and  D  B  C  equal 
to  each  other ;  then  A  B  D  and  D  B  C  will 
each  of  them  be  right  angles,  and  D  B  and 
-Q    AC  will  be  perpendicular  to  each  other. 


(g.)  An  angle  greater  than  a  right  angle  is  called  an  obtuse 
(blunt)  angle,  and  one  less  than  a  right  angle  is  called  an 
acute  (sharp)  angle. 


34  TABLES    OF    MONEY,  WEIGHTS,  ANI>    MBASCRES, 

In  Fig.  2,  D  B  A  is  an  obtuse  angle,  and  A  B  C  i?  un  acute  angle. 
In  Fig.  1,  A  B  C  is  an  acute  angle. 

(k.)  A  four-sided  figure,  having  all  its  angles  right  angles, 
is  called  a  rectangle. 

{{.)  A  rectangle,  having  all  its  sides  equal,  is  called  a 
square.  A  square,  then,  has  four  equal  sides  and  four  equal 
angles. 

Figs.  4  and  5  represent  rectangles.     Fig.  5  also  represents  a  square. 
Fig.  4.  Fig.  5. 


(y.)  If  a  square  measures  a  foot  on  each  side,  it  is  called  a 
square  foot ;  if  it  measures  a  yard  on  a  side,  it  is  called  a 
square  yard^  &c.  A  square  unit,  or  superjicial  unit,  then,  is 
any  surface  equivalent  to  a  square  1  unit  long  and  1  unit  wide. 

(k.)  All  surfaces  are  measured  by  Square  Measure,  that  is, 
by  the  number  of  squares  of  a  given  size  to  which  they  are 
equivalent.  Thus,  a  surface  contains  5  square  fee',  when  it  is 
equivalent  to  5  squares,  each  measuring  1  foot  on  a  side. 

TABLE  OF  SQUARE  MEASURE. 

144  square  inches  =  1  square  foot. 

9  square  feet  =  1  square  yard. 

304^  square  yards,  or  )  -  , 

0^70 1  i    ^  J   =  1  square  rod. 

272|-  square  feet,  ) 

40  square  rods  =  1  rood. 

4  roods  =  1  acre. 

640  acres  =  1  mile. 

^  (lA  In  this  measure,  Sq.  M.  stands  for  square  mile,  A.  for 
acre,  R.  for  rood,  sq.  rd.  for  square  rod,  sq.  yd.  for  square  yard, 
iq.  ft.  for  square  foot,  and  sq.  in.  for  square  inch. 

Note. —  Since  a  surface  a  unit  long  and  a  unit  wide  contains  a  square 


TABLES    OF    MONET,  WEIGHTS,  AND    MEABURE8.  85 

unit,  a  surface  two  units  long  and  one  unit  wide  must  contain  two  square 
units ;  a  surface  three  units  long  and  one  unit  wide  must  contain  three 
square  units ;  and  generally,  a  surface  one  unit  wide  must  contain  af 
many  square  units  as  there  are  units  in  length. 

A  surface  two  units  wide  must  contain  twice  as  many  square  units  as 
a  surface  one  unit  wide,  i.  e.,  twice  as  many  as  there  are  units  in  length ; 
a  surface  three  units  wide  must  contain  three  times  as  many  square  units 
as  a  surface  one  unit  wide,  i.  e.,  three  times  as  many  as  there  are  units 
in  length ;  and  generally,  any  surface  must  contain  as  many  square  units 
as  there  are  in  the  product  obtained  by  multiplying  its  length  by  its 
breadth.  The  surfaces  here  spoken  of  are,  in  all  cases,  supposed  to  be 
rectangular  ones. 

41 .    Cubic  Measure, 

Cubic  Measure  is  used  in  measuring  solids, 
(a.)    A  solid  is  a  magnitude  which  has  length,  breadth,  and 
thickness. 

Note.  —  The  term  "  solid,"  as  used  in  mathematics,  refers  to  space 
rather  than  to  material  substances 

{h.)  A  cube  is  a  rectangular  solid,  whose  length,  breadth, 
and  height  are  equal.  It  may  also  be  defined  as  a  solid  which 
is  bounded  by  six  equal  squares. 

(c.)  A  cube  1  foot  long,  1  foot  wide,  and  1  foot  high  would 
be  a  cubic  foot.  A  cube  1  yard  long,  1  yard  wide,  and  1  yard 
high  would  be  a  cubic  yard.  A  cubic  unit,  then,  is  any  solid 
equivalent  to  a  cube  1  unit  long,  1  unit  wide,  and  1  unit  high. 

{d.)  The  solid  contents  of  bodies  are  measured  by  cubic 
measure,  i.  e.,  by  the  number  of  cubes  of  a  given  size  which 
the  bodies  contain,  or  to  which  they  are  equivalent. 

TABLE    OP    CUBIC    MEASURE. 

1728  cubic  inches      r=  1  cubic  foot. 

27  cubic  feet  =:  1  cubic  yard, 

16  cubic  feet  =  1  cord  foot. 

8  cord  feet,  or  >  ,  _  „  , 
-» -.o  V  r  X  r  ==  1  cord  of  wood. 
128  cubic  leet,     ) 

(c.)  In  this  measure,  C.  stands  for  cord,  Cd.  ft.  for  cord 
foot,  cu.  ft.  for  cubic  foot,  cu.  yd.  for  cubic  yard,  and  cu.  in.  for 
cubic  inch.  . 


SQ  TABLES    OF    MONEY,  WEIGHTS,  AND    MEASURES. 

Note.  —  Since  a  solid  one  unit  long,  one  nnit  wide,  and  one  nnit 
high  contains  a  cubic  unit,  a  solid  one  unit  wide,  one  unit  high,  and  two 
units  long  must  contain  two  cubic  units ;  one  three  units  long  must  con- 
tain three  cubic  units  ;  one  four  units  long  must  contain  four  cubic  imits ; 
and  generally,  a  solid  one  unit  wide  and  one  unit  high  must  contain  as 
many  cubic  units  as  there  are  linear  units  in  its  length. 

Again.  Since  a  solid  two  units  wide  must  contain  twice  as  many 
cubic  units  as  a  solid  one  unit  wide,  and  a  solid  three  units  wide  must 
contain  three  times  as  many  as  a  solid  one  unit  wide,  &c.,  it  follows  that 
a  solid  one  unit  high  must  contain  as  many  cubic  units  as  there  arc  in 
the  product  of  its  length  by  its  breadth,  i.  e.,  as  many  cubic  units  as 
there  are  superficial  units  in  its  base. 

Again.  Since  a  solid  two  units  high  contains  twice  as  many  cubic 
units  as  a  solid  one  unit  high,  and  a  solid  three  units  high  contains  three 
times  as  many  cubic  units  as  a  solid  one  unit  high,  it  follows  that  any  solid 
must  contain  as  many  cubic  units  as  there  are  in  the  product  obtained 
by  multiplying  the  number  of  superficial  units  in  its  base  by  the  number 
of  linear  units  in  its  height,  i.  e.,  as  many  cubic  units  as  there  are  in  the 
product  of  its  length  multiplied  by  its  breadth,  multiplied  by  its  height. 

4:S.     Circular  or  Angular  Measure. 

(a.)  Circular  or  Angular  Measure  is  used  to  measure 
angles,  and  the  circumferences  of  circles. 

(b.)  A  circle  is  a  surface  bounded  by  a  curved  line  which 
is  every  where  equally  distant  from  a  point  within  called  the 
centre.  The  boundary  line  is  called  the  circumference  of  the 
circle. 

Eig.  6  represents  a  circle  of  which  C  is  the  centre. 

(c")  The  distance  from  the  centre  of  a 
circle  to  the  cii'cumference  is  called  the 
radius  of  the  circle. 

(c?.)    The  distance  from  a  point  on  one 

side  of  a  circle  thiough  the  centre  to  a 

point  on  the  opposite  side  is  called  the 

diameter  of  the  circle. 

{e.)   Any  portion  of  the  circumference  is  called  an  arc, 

(/.)    Every  circumference  of  a  circle,  whether  the  circle  be 

l.irge  or  small,  is  supposed  to  be  divided  into  3G0  equal  prvrts 


TABLES    OF    MONEY,  WEIGHTS,  AND    MEASURES.  87 

called  degrees.  Each  degree  is  divided  into  60  equal  parts, 
called  minutes,  and  each  minute  into  60  equal  parts,  called 
seconds. 

(g.)  A  degree  is  to  be  regarded  simply  as  the  360th  part 
of  the  circumference  of  the  circle  considered.  Hence,  it  is 
obvious  that  its  length,  and  that  of  its  subdivisions,  must  vary 
with  the  size  of  the  circle. 

(h.)  If  from  the  vertex  of  an  angle,  as  a  centre,  we  should 
draw  a  circumference,  some  portion  of  that  circumference 
would  be  included  between  the  sides  of  the  angle.  The  larger 
the  angle  is,  the  larger  will  be  the  arc  included  between  its 
sides,  and  the  smaller  the  angle  is,  the  smaller  will  be  the 
included  arc. 

(^.)    Since  the  angle  and  the  arc  thus  vary  with  each  other, 
the  arc  is  taken  as  the  measure  of  the  angle.     If  the  arc  con- 
tains 50  degrees,  the  angle  is  one  of  50  degrees,  &c. 
Fig.  7  illustrates  this. 
Fig.  7. 

(k.)  In  this  flgnre  let  C  be  the  centre  of  the 
circle  and  the  vertex  of  the  several  angles,  and 
let  F  H  and  G  D  be  lines  perpendicular  to  each 
other,  and  C  E  and  C  I  be  lines  drawn  at  random. 
Then  the  arc  E  D  is  the  measure  of  the  angle 
E  C  D  ;  the  arc  D  F,  of  the  angle  D  C  F ;  the 
arc  D  I,  of  the  angle  D  C  I ;  the  arc  E  F,  of  the 
angle  EOF,  &c. 

(Z.)  Since  F  H  and  G  D  are  perpendicular  to  each  other,  the  angles 
F  C  D,  D  C  H,  H  C  G,  and  G  C  F  are  all  right  angles,  and  each  of  them 
must  include  |.  of  the  angular  space  about  the  point  C.  Therefore  the  arc 
included  between  the  sides  of  a  right  angle  equals  i.  of  the  circumfer- 
ence, and  the  measure  of  a  right  angle  is  i  of  360  degrees,  which  equals 
90  degrees.  The  measure  of  an  acute  angle  is  less  thaii  90  degrees,  and 
that  of  an  obtuse  angle  more  than  90  degrees. 

TABLE    OF    CIRCULAR    OR   ANGULAR   MEASURE. 

60  seconds  =i  1  minute. 
60  minutes  r=  1  degree. 
860  degrees  =  1  circumference. 
(m.)    Degrees  are  marked  by  the  character  °,  minutes  by  ', 


88  TABLES    OF   MONEY,  WEIGHTS,  AND    MEASURES. 

seconds  by  ":   thus,  13°  27'  49"  =  13  degrees,  27  minutes, 
and  49  seconds. 

An  arc  of  90°  is  called  a  quadrant. 

A  sign  is  an  astronomical  measure  of  30°. 

43.    Dry  Measure, 

(a,)  Dry  Measure  is  used  for  measuring  all  kinds  of  grain, 
beans,  nuts,  salt,  &c. 

TABLE    OF   DRY   MEASURE. 

2  pints  =  1  quart. 
8  quarts  =  1  peck. 
4  pecks   =  1  bushel. 

(h.)  The  chaldron  of  36  bushels  is  sometimes  used  in 
measuring  coals,  ch.  stands  for  chaldron,  bu.  for  bushel,  pk. 
for  peck,  qt.  for  quart,  and  pt.  for  pint. 

(c.)  The  bushel  contains  2150f  cubic  inches.  The  quart 
must,  therefore,  contain  67-^  cubic  inches. 

4:4:,    Liquid  Measure, 
(a.)    All  kinds  of  liquids  are  measured  by  Liquid  Measure. 

LIQUID    MEASURE. 

4  gills     =  1  pint. 
2  pints    =  1  quart. 
4  quarts  =  1  gallon. 

(h.)  In  this  measure  gal.  stands  for  gallon,  qt.  for  quart,  pt. 
lor  pint,  and.gi.  for  gill. 

(c.)  The  hogshead  of  63  gallons  is  used  in  estimating  the 
contents  of  reservoirs  or  other  large  bodies  of  water ;  but  in 
all  other  cases,  the  term  hogshead  is  not  a  definite  measure. 
Casks  containing  from  50  or  60  to  100  or  200  gallons  are 
called  hogsheads. 

(d.)    A  barrel  of  cider  is  usually  reckoned  at  31^  gallons. 

(c.)    The  gallon  contains  231  cubic  inches. 


TABLES    OF    MONEY,  WEIGHTS,  AND    MEASURES.  39 

(/.)  The  heer  gallon  is  sometimes  used  for  measuring  milk, 
beer,  and  ale.  It  contains  282  cubic  inches.  The  beer  quart 
must  therefore  contain  70^  cubic  inches. 

4:5,    Comparison  of  Dry^  Liquid^  and  Beer  Measures. 

The  following  table  will  be  convenient  for  use  in  comparing 
Dry,  Liquid,  and  Beer  Measures  :  — 

1  qt.  Dry  Measure  =  67^  cubic  inches. 
1  qt.  Liquid  Measure  =:  57J  cubic  inches. 
1  qt.  Beer  Measure     =70^  cubic  inches. 


46. 

TABLE    0 

•F    TIME. 

60  seconds 
60  minutes 
24  hours 
7  days 

=  1  minute. 
=  1  hour. 
=  1  day. 
=  1  week. 

365;^  days,  or  ) 

52  weeks  and  1^  days  )  ^ 

(a.)  The  year  is  divided  into  12  months,  which  differ 
somewhat  in  length.  In  this  measure  yr.  stands  for  year,  mo. 
for  month,  wk.  for  week,  da.  for  day,  h.  for  hour,  m.  for 
minute,  and  sec.  for  second. 

(5.)  To  avoid  the  inconvenience  of  reckoning  :^  of  a  day 
with  each  year,  every  fourth  year  (called  leap  year)  is  reck- 
oned at  366  days,  and  the  others  are  reckoned  at  365  days. 
A  leap  year  may  always  be  known  by  this,  viz. :  Its  number 
can  be  divided  by  4.  Thus  we  know  that  1852  is  a  leap 
year,  because  1852  can  be  divided  by  4  without  a  remainder. 

(c.)  The  year  in  reality  contains  but  365  days,  5  h.  48  m. 
48  sec. ;  so  that  by  reckoning  365^^  days  we  make  a  slight  error 
each  year,  which  in  100  years  amounts  to  about  1  day.  The 
centennial  years  are  not,  therefore,  reckoned  as  leap  years, 
unless  the  number  of  the  year  be  divisible  by  400.  Thus  the 
year  1900  wiU  not  be  a  leap  year ;  but  the  year  2000  will  be. 


40  NOTATION   AND    NUMERATION. 


TABLE    OP    THE    MONTHS. 


January  has  31  days. 
February  *  has  28  days. 
March  has  31  days. 
April  has  30  days. 
May  has  31  days. 
June  has  30  days. 


July  has  31  days. 
August  has  31  days. 
September  has  30  days. 
October  has  31  days. 
November  has  30  days. 
December  has  31  days. 


^LT".    Miscellaneous, 

12  things  =  1  dozen. 
12  dozen  =  1  gross. 
12  gross   =  1  great 
20  things  z=  1  score. 

A  barrel  of  beef  or  pork  weighs  200  lbs. 
A  barrel  of  flour  weighs  196  lbs. 

PAPER. 

24  sheets  =.  1  quire. 
20  quires  =  1  ream. 

BOOKS. 

A  sheet  folded  into    2  leaves  is  called  a  folio. 
u     a         u         a      4      «      «      «      «  quarto. 
«     "         a         «      3      «      «      «      u  octavo. 
«     «         «         a     12      «      "      «      "  duodecimo  or  12rao. 
II     u         u        a    18      "      "      "      "  18mo. 

4:8.    French  Pleasures  and  Weights, 

(a.)    The  following  measures  and  weights  are  often  referred  to  in  this 
country,  especially  in  scientific  works. 

FSEXCH   LONG   MEASURE. 

10  millimetres  =  1  centimetre. 

10  centimetres  =  1  decimetre. 

10  decimetres  =  1  metre. 

10  metres  =  1  decametre. 

10  decametres  =  1  hectometre. 

•  In  leap  year  February  hais  29  days. 


ADDITION.  41 

10  hectometres  =  1  kilometre. 
10  kilometres     =  1  myriametre. 

The  metre  is  regarded  as  the  unit  of  measure,  and  equals  39.371  of 
our  inches.  It  is  the  twenty-miilionth  part  of  the  distance  measured  on 
a  meridian,  from  one  pole  to  the  other. 

(6.)     FEENCH   WEIGHTS. 

10  milligrammes  =  1  centigramme. 

10  centigrammes  =  1  decigramme. 

10  decigrammes  =  1  gramme. 

10  grammes  =  1  decagramme. 

10  decagrammes  =  1  hectogramme. 

10  hectogrammes  =  1  kilogramme. 

10  kilogrammes  =  1  myriagramme. 

The  gramme  is  regarded  as  the  unit  of  this  weight,  and  equals  15.434 
Troy  grains. 

The  kilogramme  is  the  weight  most  frequently  used  in  business  trans- 
actions, and  equals  15434  Troy  grains,  or  very  nearly  2.204857  pound* 
Avoirdupois. 

(C.)    FRENCH  MONEY. 

10  centimes  =  1  decime. 
10  decimes  =  1  franc. 

The  franc  =  $.186;  hence,  the  five-franc  piece,  often  seen  in  the 
United  States,  is  equal  in  value  to  93  cents. 


SECTION  IV. 

ADDITION. 

40,    Definitions^  Illustrations ^  and  Explanations, 
(a.)    Addition   is   the   process   by   which,   having 

ftfiVERAL    NUMBERS    GIVEN,   WE    FIND    A   NUMBER    EQUAL    IN 
VALUE    to    all    op    THEM. 

(b.)     The    number  thus   obtained   is   called   the    sum   or 

AMOUNT. 


42  ADDITION. 

Thus  :  How  many  are  7  -f-  3  +  5,  would  be  a  problem  in  addition, 
and  the  answer,  15,  would  be  the  sum  of  7,  3,  and  5. 

(c.)  In  order  that  numbers  may  be  added,  it  is  necessary 
that  the  things  they  represent  shall  be  of  the  same  name  or 
denomination. 

Illustrations.  —  2  books  and  3  slates  would  be  neither  5  books  nor 
5  slates ;  but  since  books  and  slates  are  both  things,  we  can  change  the 
denomination  of  both  by  calling  them  things ;  when  we  shall  have,  2 
things  and  3  things  are  5  things. 

In  like  manner  2  tens  and  3  units  would  be  neither  5  tens  nor  5 
units  ;  but  by  reducing  the  tens  to  units,  calling  them  20  units,  we  shall 
have  2  tens  -}-  3  units  =  20  units  -j-  3  units  =  23  units. 

2  shillings  -J-  3  pence  are  neither  5  shillings  nor  5  pence;  but  since  2 
shillings  =  24  pence,  2  shillings  -f-  3  pence  must  equal  24  pence  -f-  3 
pence,  or  27  pence. 

(d.)  For  such  reasons  it  will  be  found  convenient,  in  writing 
large  numbers  for  addition,  to  write  those  of  the  same  denom- 
ination near  each  other.  This  can  best  be  done  by  writing 
them  in  vertical  columns,  so  that  units  shall  come  under  units, 
tens  under  tens,  &c.,  and  pounds  under  pounds,  shillings  under 
shillings,  pence  under  pence,  &c. 

(e.)  In  adding,  we  can  Ibegin  with  any  denominatioii  we 
choose ;  but  it  will  usually  be  more  convenient  to  begin  with  the 
lowest,  or  the  one  at  the  right  hand,  and  to  reduce  the  sum  of 
each  column  to  a  higher  denomination,  when  it  can  be  done. 

Note.  —  Addition  is  the  most  important  of  the  four  numerical  opera- 
tions, both  because  it  is  the  foundation  of  all  the  others,  and  because  it  is 
the  one  most  frequently  used  in  all  the  departments  of  practical  life. 
Moreover,  it  is  the  one  in  which  there  is  the  greatest  liability  to  error. 
For  these  reasons,  and  many  others  which  might  be  urged,  the  student 
should  be  very  careful  to  master  it  fully. 

(f.)  The  methods  of  applying  these  principles  are  illustrat- 
ed in  the  following  examples  and  solutions. 

50.    Simple  Addition. 

(a.)  Abstract  numbers,  or  concrete  numbers,  which  repre- 
sent values  in  terms  of  a  single  denomination,  as  in  pounds^ 
in  bushels,  or  in  dollars,  are  called  simple  numbers  ;  but 
concrete  numbers,  which  represent  values  in  terms  of  several 


ADDITION.  48 

different  denominations,  as  in  pounds^  shillings,  and  pence,  or 
in  bushels,  pecks,  and  quarts,  are  called  compound  numbers. 

(h.)    Simple  Addition  is  the  addition  of  simple  numbers. 

What  is  the  sum  of  75798  +  24687  +  39764  +  86328  + 
4395  +  283  +  86536  ? 

Solution.  —  We  first  write  the  numbers,  placing  units  under  units, 
tens  under  tens,  &c.,  in  order  that  figures  expressing  the  same  denomi 
nations  may  be  near  together     Thus  :  — 

75798. 
24687. 
39764. 
86328. 
4395. 
283. 
86536. 


317,791. 

*  Beginning  at  the  bottom  of  the  units  column,  (because  the  lowest 
denomination  mentioned  is  units,)  and  naming  only  results,  we  add  thus : 
6,  9,  14,  22,  26,  33,  41  units,  which  are  equal  to  4  tens  and  1  unit. 

Writing  1  as  the  units'  figure  of  the  amount,  we  add  the  4  tens  with 
the  figures  of  the  tens  column  j  thus,  4,  7,  15,  24,  26,  32.  40,  49  tens, 
which  are  equal  to  4  hundreds  and  9  tens. 

Writing  9  as  the  tens'  figure  of  the  amount,  we  add  the  4  hundreds 
with  the  figures  of  the  hundreds  column,  thus  ;  4,  9,  11,  14,  17,  24,  30, 
37  hundreds,  which  are  equal  to  3  thousands  and  7  hundreds. 

Writing  7  as  the  hundreds'  figure  of  the  amount,  we  add  the  3  thou- 
sands with  the  figures  of  the  thousands  column  ;  thus,  .",  9,  13,  19,  28, 
32,  37  thousands,  which  are  equal  to  3  ten-thousands  and  7  thousands. 

Writing  7  as  the  thousands'  figure  of  the  amount,  we  add  3  ten-thou 
sands  with  the  figure  of  the  ten-thousands  column;  thus,  3,  11,  19.  22, 
24,  31  ten-thousands,  which  are  equal  to  3  hundreds  thousands  and  1 
ten-thousand ;  and  as  there  are  no  higher  denominations,  we  write  the  3 
and  1  in  their  appropriate  places. 

Having  thus  added  all  the  denominations,  we  must  have  the  sum,  or 
amount  of  the  numbers,  which  is  317,791. 

Note.  —  Many  call  the  names  of  the  separate  numbers  added,  as 
well  as  the  results  of  the   addition,  and  would   add,  thus :    6  and   3 

*  If  the  learner  does  not  readily  understand  this  method  of  addition, 
let  him  for  a  time  call  the  separate  numbers  added,  as  explained  in  the  note. 


44  ADDITION. 

are  9,  and  5  are  14,  and  8  arc  22,  and  4  are  26,  and  7  are  33,  aijd  8  ar« 

41  ;  41  units  are  equal,  &c.  This  method  is  much  less  expeditious  than 
the  first  one  given,  and  therefore  should  not  be  adopted.  Indeed,  we 
may  say,  that  calling  the  names  of  the  separate  numbers  comprising  a 
sum  is  to  addition  what  spelling,  i.  e.,  calling  the  letters  composing  a 
word,  is  to  reading. 

P  51.     Compound  Addition, 

(a.)  Compound  Addition  is  the  addition  of  compound 
numbers. 

Wliat  is  the  sum  of  £8  15  s.  11  d.  2  qr.  +  £3  12  s.  8  d.  3  qr. 
+  £9  19s.  9d.  Iqr.  +  £7  18s.  lOd.  3qr. +  £5  18s.  2qr. 
+  £6  8d.  Iqr. +  £5  13s.  3d.  +  16s.  8d.  Iqr.? 

Solution. — "We  first  write  the  numbers,  placing  pounds  under  pounds, 
shillings  under  shillings,  &c.,  in  order  that  figures  expressing  the  same 
denomination  may  be  near  each  other. 


£ 

s. 

d. 

qr. 

8 

15 

11 

2 

3 

12 

8 

3 

9 

19 

9 

1 

7 

18 

10 

3 

5 

18 

0 

2 

6 

0 

8 

1 

5 

13 

3 

0 

16 

8 

1 

48     16       0     1  =  Amount. 

Beginning  with  the  right  hand  column,  as  before,  and  reducing  as  we 
add,  we  proceed  thus  :      1  qr.  and  1  qr.  are  2  qr.,  and  2  qr.  are  4  qr.  = 

1  d.,  and  3  qr.  are  1  d.  3  qr.,  and  1  qr.  are  1  d.  4  qr.  =  2  d.,  and  3  qr.  are 

2  d.  3  qr.,  and  2  qr.  are  2  d.  5  qr.  =  3  d.  1  qr. 

"Writing  1  as  the  farthings'  figure  of  the  sum,  we  add  the  3  d.  with 
the  numbers  in  the  pence  column,  thus:    3  d.  and  8  d.  are  11  d.,  and 

3  d.  are  14  d.  =  1  s.  2  d.,*  and  8  d.  are  1  s.  10  d.,  and  10  d.  are  1  s.  20  d. 
=  2  s.  8  d.,*  and  9  d.  ai-e  2  s.  17  d.  =  3  s.  5  d.,*  and  8  d.  are  3  s.  13  d. 
=  4  s.  1  d.,*  and  lid.  are  4  s.  12  d.*  =  5  s. 

"Writing  0  as  the  pence  figure  of  the  sura,  we  add  the  5  s.  with  the 
numbers  in  the  shillings  column,  thus:    5  s.  and  16  s.  are  21  s.  =  £1 


•  Since  12  d.  =  1  s 


ADDITION.  45 

I  s.,*  and  13  s  =  £1  14  s.,  and  18  s.  are  £1  32  s.*  =  £2  12  s.,  and  18  s. 
are  £2  30  s.  =  £3  10  s.  *  and  19  s.  are  £3  29  s.  =  £4  9  s.  *  and  12  s. 
are  £4  21  s.  =  £5  1  s.,=*  and  15  s.  are  £5  16  s. 

Writing  16  as  the  shillings'  figure  of  the  sum,  we  add  £5  with  the 
numbers  of  the  pounds  column,  thus:  5,  10,  16,  21,  28,  37,  40,  48. 

As  all  the  denominations  of  the  given  numbers  have  been  added,  the 
amount  sought  must  be  £48  16  s.  0  d.  1  qr. 

(b.)  In  the  above  form  of  solution,  the  numbers  to  be 
added  have  been  named  merely  to  insure  that  the  explana- 
tions should  be  understood,  but  in  practical  work  the  addition 
should  be  performed  by  naming  only  results. 

Thus  :  1  qr.,  2  qr.,  4  qr.  =  1  d. ;  1  d.  3  qr..  Id.  4  qr.  =  2  d. ;  2  d. 
3  qr.,  2d.  5  qr.  =  3  d.  1  qr.  Write  1  qr.  3d.,  11  d.,  14  d.  =  1  s.  2  d. ;  Is. 
10  d.,  1  s.  20  d.  =  2  s.  8  d. :  2  s.  17  d.  =  3  s.  5  d.,  &c. 


53.    Compound  and  Simple  Addition  compared. 

(a.)  Compound  Addition  involves  precisely  the  same  prin- 
ciples that  Simple  Addition  does.  In  both,  numbers  of  the  same 
denomination  are  placed  under  each  other,  in  order  that  they 
may  be  more  readily  distinguished.  In  both,  we  commence 
to  add  at  the  lowest  denomination,  in  order  to  avoid  the 
necessity  of  erasing  or  altering  figures  which  have  been  once 
written ;  in  both,  we  reduce  the  sum  of  each  column  to  units 
of  the  next  higher  denomination,  in  order  that  the  answer 
may  appear  in  its  simplest  form ;  and  in  both  we  add  the 
units  thus  obtained  with  those  written  in  the  column  of  the 
next  higher  denomination.  Moreover,  the  same  methods  of 
proof  apply  to  both. 

(b.)  The  slight  differences  in  the  methods  of  applying  these 
principles  result  from  the  fact,  that  in  simple  numbers  10 
units  of  one  denomination  always  equal  one  of  the  next 
higher,  while  in  compound  numbers  there  is  no  uniformity  in 
this  respect. 

Note  to  the  Teacher.  —  The  explanations  and  examples  are  so 
arranged  that,  should  the  teacher  think  it  inexpedient  to  teach  Com- 
pound Addition  at  the  same  time  that  Simple  Addition  is  taught,  he  can 

♦  Since  20  s.  =  £1. 


46  ADDITION. 

defer  it  till,  in  his  opinion,  the  classjare  prepared  for  it.  We  would, 
however,  recommend  that  whenever  it  is  taught,  it  should  be  presented 
as  a  further  application  of  the  principles  involved  in  Simple  Addition. 

The  same  thing  may  be  said  of  Simple  and  Compound  Subtraction, 
Multiplication,  and  Division. 

53.    Methods  of  Proof. 

(a.)  We  can  test  the  correctness  of  the  work  in  many 
ways,  a  few  of  which  we  will  mention. 

First  Method.  —  Go  over  the  work  carefully  a  second  time 
in  the  same  manner  as  at  first. 

Second  Method.  —  Begin  to  add  at  a  different  part  of  the 
column  from  that  at  which  the  first  addition  was  commenced ; 
i.  e.,  if  the  first  addition  was  commenced  at  the  top,  begin  the 
second  at  the  bottom,  and  vice  versa.  This,  by  presenting 
the  figures  in  a  different  order,  renders  it  improbable  that  any 
mistake  which  may  have  been  made  in  the  first  work  will  be 
repeated. 

Third  Method.  —  Separate  the  numbers  to  be  added  into 
two  or  more  parts,  add  the  parts  separately,  and  then  adi 
their  sums.     This  method  is  illustrated  below. 

PROOF    OF    EXAMPLE    IN   50    BY   THIRD    METHOD. 

75798 


24687 
89764 
86328 

43951 

283 

86536  J 


■  1st  part* 


■  2d  part. 


317791  =  First  Answer. 


a  =  226577  =  Sum  of  first  part. 
b=    91214=    "      «   second  part. 

317791  =  Sum  of  the  two  partial  sums  =  first  a  ♦'^•^r 
Thus  showing  that  the  work  was  correct. 


ADDITION. 


47 


PROOF    OF    EXAMPLE    IN   ^1    BY   THIRD     METHOD. 

£.     s.      d.    qr. 
8     15     11     2  " 

'  1st  part. 


'  2d  part. 


3 

12 

8 

3 

9 

19 

9 

1 

7 

18 

10 

3 

5 

18 

0 

2 

5 

0 

8 

1 

5 

13 

3 

0 

16 

8 

1 

48     16       0     1=:  First  Answer. 


30       7       4     1  =  Sum  of  1st  part. 
18       8       8     0=  Sura  of  2d  part. 


48  16  0  1  =  Sum  of  partial  sums  r=  first  answer  ob- 
tained.    Thus  showing  the  work  to  be  correct. 

Note.  —  When  long  columns  are  to  be  added,  it  may  sometimes  be 
convenient  to  divide  them  in  this  way  in  performing  the  first  addition. 
The  student  should,  however,  accustom  himself  to  adding  the  longest 
columns  without  any  separation  into  parts. 

Fourth  Method.  —  Beginning  either  at  the  right  or  at  the 
left  hand  to  add,  write  the  sum  of  each  denomination  sepa- 
rately, and  then  add  these  sums  together. 

Fifth  Method.  —  Begin  with  the  left  hand  column,  and  pro- 
ceed as  follows :  — 

PROOF   OF   THE   EXAMPLE   IN    50. 

By  adding  the  ten-thousands  column  we  find  that  its  sum  is  28; 
but  as  there  are  31  ten-thousands  in  the  answer  first  obtained,  we  infer 
that  3  ten-thousands  were  brought  from  the  lower  denominations.  3 
ten-thousands  =  30  thousands,  which,  added  to  the  7  written  in  the 
thousands'  place  of  the  answer,  gives  37  thousands  to  be  accounted  for. 
The  sum  of  the  thousands  column  is  34,  which,  taken  from  37,  leaves 
3;  thus  showing,  that  if  the  work  is  correct,  3  thousands  must  have 
been  brought  from  the  lower  denominations.  3  thousands  ^=^  30  him- 
dreds,  which,  added  to  the  7  written  in  the  hundreds'  place  of  the  answer, 


•48  ADDITION. 

gives  37  hundreds  to  be  accounted  for.  The  sum  of  the  hundreds 
column  is  33,  which,  taken  from  37,  leaves  4  ;  thus  showing,  that  if  the 
work  is  correct,  4  hundreds  must  have  been  brought  from  the  lowei 
denomination.  4  hundreds  =  40  tens,  which,  added  to  the  9  tens  writ- 
ten in  the  tens'  place  of  the  answer,  gives  49  tens  to  be  accounted  for 
The  sum  of  the  tens  c6lumn  is  45,  which,  taken  from  49,  leaves  4  ;  thus 
showing,  that  if  the  work  is  correct,  4  tens  must  have  been  brought 
from  the  units  column.  4  tens  =  40  units,  which,  added  to  the  1  unit 
written  in  the  units'  place  of  the  answer,  gives  41  units  to  be  accoun^ 
ed  for. 

As  the  sum  of  the  units  column  is  41,  we  infer  that  the  work  ia 
correct. 

PROOF    OF    THE    EXAMPLE   IX    51. 

By  adding  the  pounds  column,  we  find  its  sum  is  £43,  which,  taken 
from  the  £48  written  in  the  answer,  leaves  £5  ;  thus  showing,  that  if 
the  answer  is  correct,  £5  must  have  been  brought  from  the  lower  de- 
nominations. £5=  100  s.,  which,  added  to  the  16  s.  written  in  the 
answer,  gives  116  s.  to  be  accounted  for.  The  sum  of  the  shillings 
column  is  111  s.,  which,  subtracted  from  the  116  s.,  leaves  5  8.;  thus 
showing,  that  if  the  answer  is  correct,  5  s.  must  have  been  brought  from 
the  lower  denominations.  5  s.  =  60d.,  which,  as  there  are  no  pence 
written  in  the  answer,  gives  60  d.  to  be  accounted  for. 

The  sum  of  the  pence  column  is  57  d.,  which,  taken  from  60  d.,  leaves 
3  d. ;  thus  showing,  that  if  the  answer  is  correct,  3  d.  must  have  been 
brought  from  the  column  of  farthings ;  3  d.  =  12  qr.,  which,  added  to 
the  1  qr.  written  in  the  answer,  gives  13  qr.  to  be  accounted  for. 

As  the  sum  of  the  farthings  column  is  13,  we  infer  that  the  answer  is 
corre-et 

(h.)  The  first,  second,  and  third  methods  of  proof  are  the 
most  practical,  but  as  the  fourth  and  fifth  furnish  valuable  illus- 
trations of  the  nature  of  the  various  changes  and  reductions, 
and  call  the  reasoning  faculties  into  healthful  exercise,  they 
should  not  be  omitted  by  the  student. 

(c.)  If,  by  any  of  these  methods,  we  obtain  a  different  result 
from  the  one  we  first  obtained,  we  may  be  sure  there  is  an 
error  in  one  or  both  operations,  and  should  examine  both  care- 
fully to  find  it 

(d.)  Some  method  of  proof  should  always  be  resorted  to, 
until  the  pupil  aequires  sufficient  skill  to  be  sure  of  the  accu- 
racy of  his  work  without  it. 


ADDITION.  49 


54.    Importance  of  Accuracy  and  Certainty. 

(a.)  No  person  who  is  willing  to  allow  an  error  to  pass  undetect- 
ed can  be  a  good  arithmetician.  Accuracy^  absolute  accuracy,  should 
be  aimed  at  in  every  operation  ;  and  no  labor  is  too  great  which  is 
necessary  to  secure  it.  Not  only  should  the  results  be  accurate,  but  the 
computer  should  know  for  himself  that  they  are  so.  If  he  has  any  doubt 
concerning  a  result,  he  should  examine  each  and  every  step  of  his  work, 
to  see, 

First.  TJiat  it  was  a  proper  one  to  take. 

Second.  That  it  was  taken  at  the  right  time. 

Third.  That  it  was  taken  correctly. 

(b.)  One  problem  thus  solved  and  proved  by  a  learner  is  of  more  real 
value  to  him  than  ten  solved  by  him  and  proved  by  another,  or  tested 
by  comparison  with  a  printed  answer.  The  accountant  does  not  hesi- 
tate to  spend  hours,  and  even  days,  in  looking  over  long  and  complicated 
accounts,  to  discover  the  cause  of  an  error  of  a  few  cents  in  a  trial 
balance  sheet,*  and  surely  the  student  ought  not  to  shrink  from  the  task 
of  proving  the  correctness  of  his  solutions  of  the  much  more  simple 
problems  contained  in  a  school  text  book. 

(c.)  Rapidity  in  the  performance  of  numerical  operations  is  scarcely 
of  secondary  importance  to  accuracy  and  certainty.  The  most  accurate 
computers  are  usually  the  most  rapid  in  their  work. 

*  The  trial  balance  sheet  is  used  in  keeping  books  by  double  entry,  as 
a  means  of  determining  whether  any  errors  exist  in  the  entries  which  have 
been  made  in  some  given  time,  as  a  month,  a  quarter,  (i.  e.,  three  months,) 
six  months,  or  a  year.  By  its  aid,  the  existejice  of  an  error  may  be  ascer- 
tained ;  but  the  error  itself  cannot  be  discovered  without  examining  the 
separate  entries  and  accounts. 

An  intelligent  and  highly  accomplished  accountant,  who  has  charge  of 
the  books  of  a  large  manufacturing  establishment,  employing  three  hundred 
men,  once  spent  nearly  a  week  in  examining  his  accounts,  to  discover  the 
cause  of  an  error  of  a  ffew  cents ;  and  said  he,  **  I  never  spent  the  same 
amount  of  time  more  profitably."  Another  gentleman,  bearing  also  a  high 
reputation,  and  receiving  a  good  salary  as  an  accountant,  spent,  to  use  his 
own  language,  "the  greater  part  of  four  days  in  searching  out  the  cause 
of  an  error  of  ten  cents."  Both  these  gentlemen  say,  that  if  they  should 
adopt  any  other  principle  than  that  of  absolute  accuracy,  they  could  not 
retain  their  situations.  Every  accountant,  business  man,  and  practical 
man  bears  similar  testimony,  and  confirms  these  views.  Indeed,  most  of 
them  say,  that  the  knowledge  of  arithmetic  acquired  in  the  school  room 
has  been  of  little  practical  value  to  them,  because  they  did  not  learn  to  be 
accurate  and  rapid  in  performing  their  work,  and  to  know  for  themselves 
that  they  had  been  accurate. 
5 


*Q 


ADDITION. 


5S^  Problems  for  Solution. 
Add  the  numbers  in  the  following  examples  :  — 

1.  2.                           3. 

476392  832547       76486795 

584273  482938       37639487 

143253  548276       15327943 

624415  398254       82753428 


4. 

5. 

6. 

206805795 

5786.73 

379.46 

740376525 

2974.37 

82.375 

814256324 

8362.32 

986.3 

234567890 

4593.67 

59.486 

325462813 

5876.79 

576.829 

453269983 

2394.16 

403.586 

7. 

8. 

9. 

4273.75 

43.057 

786.73 

28.9873 

2.6875 

2.7598 

5794.0000 

138.2654 

.58678 

38.5946 

53.4867 

6.82594 

3954.2765 

867.9583 

36.95006 

5.7986 

15.8787 

.00487 

.385 

587.483 

30.2857 

59.8678 

865.9486 

5.7642 

10. 

H. 

12. 

8732.175 

8.7037 

49.0064 

6243.262 

2.9675 

2.206 

8711.547 

86.08972 

69.0425 

2952.364 

5.86102 

87.63214 

428.249 

2.407967 

83.8006 

1497.168 

82.27313 

70.3728 

6557.438 

25.7529 

74.857632 

4386.295 

4.8063 

428.4269 

ADDITION.  51 

13.  What  is  the  sum  of  679487  +  386754  +  329687  + 
435429  +  276834  +  579487  ? 

14.  What  is  the  sum  of  324067  +  235143  +  543345  + 
425341  +  876583  +  947869  ? 

15  What  is  the  sum  of  $473.87  +  $526.94  +  $857.93 
+  $297.16  +  $87.43  +  $528.60  +  $35.29  ? 

16.  What  is  the  sum  of  857436.57  +  25986.483  + 
295463.867  +  297484.253  +  80672.005  ? 

17.  What  is  the  sum  of  258.647943  +  547.685329  + 
27.84372  +  9765.4837  +  736.852066  +  542.063794  ? 

18.  What  is  the  sum  of  984137.612  +  257.00684  + 
43687.5792  +  574869.23757  +  2068439.14238+  1748.2 
+  13.37  ? 

19.  AYhat  is  the  sum  of  1864  +  437.29  +  58.697  + 
12.86  +  7527.385  +  167.97  +  848.396  +  4.584? 

20.  What  is  the  sum  of  389.40067  +  2768.4372  + 
5894.276  +  1385.7281  ? 

21.  What  is  the  sum  of  728  +  436  +  549  +  278  +  367 
+  825? 

22.  Wliat  is  the  sum  of  426764572681  +  894737629437 
+  179428630006  +  576428670639  +  584967245876? 

23.  What  is  the  sum  of  3798643  +  5978642  +  5489379 
+  675986  +  3768543  +  27864  +  3798742  +  8957387  + 
9583796  +  8395989  +  3865372  ? 

24.  What  is  the  sum  of  83679  +  54873  +  72352  + 
95873  +  8756  +  35906  +  87506  +  29764  +  38756  + 
35742  ? 

25.  What  is  the  sum  of  57386  +  2864.3  +  379.86  + 
28.697  +  5.4738  +  .97986  +  7.5983  +  86.794  +  886.79 
+  2937.6  +  70003  +  9764.2  +  859.86  +  48.375  ? 

26.  What  is  the  sum  of  $8.69  +  $13.48  +  $4.48  + 
$8.64  +  $37.15  +  $47.13  +  $.86  +  $.25  +  $9.37  + 
$6.08  +  $3.54  +  $7.06  +  $2.37  +  $4.68  +  $20.08  + 
$7.57  +  $7.48  ? 

27.  What  is  the  sum  of  $4,175  +  $3,867  +  $5,384  + 
$9,375  +  $5.78  +  $8,378  +  $2,635  +  $.875  +  $1.25  + 


9^  ADDITION. 

S4.28  +  $3.19  -1-  $8,625  +  $5,846  +  $9,738  +  $5.96  -|- 
$7.50  4-  $3.25  ? 

28.  What  is  the  sum  of  $.27  -f  $.63  +  $1.04  +  $.50  + 
S.375  +  $1.50  +  $.07  +  $.42  +  $.625  -f  $.375  +  $3.27 
4-  $5.94  +  $.86  4-  $1.83  +  $.06  +  $.40  +  $.125  -|. 
$1.33? 

29.  What  is  the  sum  of  $85.76  +  $77.25  +  $.86  -[.. 
$34.50  +  $7.38  +  $50.50  +  $7.13  +  $.47  +  $-^8  -f 
$28.17  -f  $29.50  +  $8.07  +  $5.00  +  $17.84  +  $.03  + 
$5.28  ? 

30.  What  is  tlie  sum  of  58694  +  67867.9432  +  45879.- 
8376  +  28697.4  +  38679.58432  +  27598.542  +  36789.754 
+  58767.5437  +  86427.58697  +  98003.79  +  28547.3298 
+  28475.9767  ? 

31.  W^hat  is  the  sum  of  958679.4437  +  298673.925  + 
586732  +  9678.4593  +  486.7923  +  5878.6532  -|-  185.3 
+  28.6734  +  86.79635  +  28-76  +  59,836  +  45173.425  ? 


32. 

33. 

£.       s.       d. 

qr. 

£. 

s. 

d.  qr. 

17  13  11 

1 

164 

13 

8   1 

18  15   8 

3 

231 

0 

6  3 

29  19   6 

1 

485 

19 

11  2 

47   8  10 

3 

738 

18 

2  3 

25  13  11 

2 

487 

16 

10  0 

87  18   9 

3 

833 

19 

11  3 

34. 

35. 

T.  cwt.  qr.  lb. 

oz. 

T 

cwt,  qr. 

lb. 

oz.  dr. 

13  18  2  23 

14 

8 

4  3 

7 

5   9 

23  19  1  24 

15 

2 

18  1 

24 

15  15 

6   8  0  17 

3 

9 

19  3 

24 

15  15 

24  16  1  24 

12 

7 

13  0 

20 

11  14 

3  19  0  20 

9 

6 

17  3 

15 

10  13 

8  13  3  22 

13 

7 

14  1 

22 

11  12 

ADDITION, 

5S 

36. 

37. 

lb. 

oz. 

dwt 

.    gr. 

lb.    oz 

.    dwt. 

gr- 

4 

6 

18 

23 

5       9 

1     19 

21 

2 

11 

13 

5 

2     11 

13 

23 

1 

7 

3 

21 

1       ^ 

;     6 

12 

8 

16 

20 

2       5 

1       9 

6 

1 

10 

9 

17 

3     10 

»     19 

19 

5 

9 

18 

20 

8       7 

17 

22 

38 

39 

.* 

lb 

S      5 

9 

gr- 

m. 

fur.    rd. 

yd.    ft. 

in. 

6 

8     5 

2 

16 

7 

6     37 

4     2 

8 

9 

8     4 

1 

17 

9 

4     24 

2     1 

2 

3 

9     2 

2 

8 

6 

7     34 

1     2 

3 

7 

3     7 

1 

19 

8 

5     28 

1     1 

7 

4 

6     6 

2 

13 

9 

3     37 

2     2 

8 

9 

8     3 

2 

15 

7 

4     19 

1     1 

6 

40.t 

414 

m.  ftir. 

rd. 

yd. 

ft.     iiL 

rd.   vd.    ft. 

in. 

8     6 

34 

4 

2      7 

8 

2     1 

11 

7     2 

38 

3 

1     11 

3 

5     2 

6 

4    5 

12 

5 

2       4 

4 

3     1 

9 

2     7 

26 

4 

1     10 

7 

2     0 

8 

9     3 

32 

5 

0       9 

3 

4     2 

10 

8     5 

13 

3 

2       6 

7 

2     2 

3 

*  In  adding  yards,  it  will  usually  be  well  to  consider  every  eleven  yards 
as  two  rods.  Such  a  course  will^  to  a  very  great  extent,  avoid  the  use  of 
fractions.  The  pupil  should,  however,  bear  in  mind  that  half  of  a  yard  =c 
1  ft.  6  in.,  and  that  when  in  any  number  there  are  5  yards,  and  1ft.  6  in. 
besides,  the  value  may  be  better  expressed  as  1  rod. 

t  Obtaining  the  answer  to  the  40th  example  in  the  usual  method,  we 
shall  find  it  to  be  41  m.  7  fur.  39  rd.  fi  yd.  2  ft.  11  in.  This  is  correct,  but  it 
is  not  in  the  best  form,  for  although  there  are  not  units  enough  expressed 
©f  any  denomination  to  make  one  of  the  next  higher,  it  equals  42  m.  0  fur. 
0  rd.  0  yd.  1  ft.  5  in.  Show  the  truth  of  this  statement,  and  show  also  tehy 
the  answer  does  not  at  first  appear  in  the  best  form. 

X  The  answer  to  the  41st  example  will  take  the  form  at  II  st  of  35  rd 
5* 


54 


ADDITION. 

42. 

43.* 

m. 

fur. 

rd.  yd. 

ft. 

in. 

A. 

R. 

?3: 

Td. 

f?- 

in. 

3 

4 

19  3 

1 

6 

16 

2 

28 

19 

7 

132 

8 

2 

16  5 

0 

5 

14 

3 

37 

8 

2 

47 

9 

4 

39  4 

2 

9 

66 

1 

19 

17 

3 

142 

2 

1 

25  3 

2 

11 

18 

2 

39 

16 

4 

27 

5 

2 

38  5 

1 

3 

17 

3 

21 

28 

8 

99 

8 

7 

18  4 

2 

10 

25 

2 

31 

30 

6 

100 

50.     Another,   and   often    a    shorter,   Method   of  reducing 
Compound  Numbers. 

It  will  often  be  more  convenient  to  make  the  reductions  by 
adding  enough  of  one  number  to  another  to  give  a  sum  equiv- 
alent to  a  unit  of  the  next  higher  denomination.  We  will 
take,  for  illustration,  the  example  given  in  51, 


JL 

s. 

d. 

qr. 

8 

15 

11 

2 

3 

12 

8 

3 

9 

19 

9 

1 

7 

18 

10 

3 

5 

18 

0 

2 

6 

0 

8 

1 

5 

13 

3 

0 

16 

8 

1 

48     16       0     1 

Explanation.  —  Having  found  the  sum  of  the  farthings 
column  to  be  equal  to  3  d.,  we  add  the  3  d.  with  the  column 
of  pence,  thus :  — 


4Jyd.  2  ft.  11  in  ,  which  should  be  changed,  for  the  sake  of  simplicity,  to 
35  rd.  5  yd.  1  ft,  5  in.  Show  the  equality  of  the  two  expressions,  and  the 
method  by  which  the  reduction  can  be  made. 

*  In  adding  square  yards,  it  will  be  of  service  to  notice  that  60J  sq.  yd, 
=  2  sq.  rd  ;  that  90|  sq.  yd.  =  3  sq.  rd. ;  that  121  sq.  yd.  =  4  sq.  rd ;  and 
that  4  of  a  sq.  yd.  =  2  sq.  ft.  36  sq.  in,  This  will  avoid  any  difficulty  in  th» 
use  of  fractions. 


ADDITION.  55 

3  d.  and  8  d.  are  II  d.,  and  3  d.  are  1  s.  2  d.,  (by  adding  1  of  the  3  d. 
with  the  11  d.,)  and  8  d.  are  1  s.  10  d.,  and  10  d.  are  2  s.  8  d.,  (by  adding 
2  of  one  10  d.  with  the  other  10  d.,)  and  9  d.  are  3  s.  5  d.,  (by  adding  3  of 
the  8  d.  with  the  9  d ,)  and  8  d.  are  4  s.  1  d.,  (by  adding  4  of  the  5  d.  with 
the  8  d  ,)  and  1 1  d.  are  5  s.  0  d. 

Adding  the  5  s.  with  the  shillings  column,  we  have  5  s.  and  16  s.  are 
£1  I  s.,  (by  adding  4  of  the  5  s.  with  16  s.,)  and  13  s.  are  £1  14  s.,  and 
'  18  s.  are  £2  12  s.,  (by  adding  2  of  the  14s.  with  18s.,)  and  18  s.  are 
£3  10  s.,  (by  adding  2  of  the  12  s.  with  18  3.,)  and  19  s.  are  £4  9  s.,  (by 
adding  1  of  the  10  s.  with  the  19  s.,)  and  12  s.  are  £o  1  s.,  (by  adding  8 
of  the  9  s.  with  12  s.,)  and  15  s.  are  £5  16  s.  =  sum  of  shillings  column. 

(b.)  We  have  mentioned  the  numbers  added  in  order  to 
secure  clearness  of  explanation,  but  in  practical  woit  the 
results  alone  should  be  named. 

Thus,  beginning  with  the  farthings,  we  have,  — 

1  qr.,  2  qr.,  1  d.,  1  d.  3  qr.,  2  d.,  2  d.  3  qr.,  3  d.  1  qr.     Write  1  qr. 

3d.,  11  d.,  1  s.  2  d.,  I  s.  10  d.,  2  s.  8  d.,  3  s.  5  d.,  4  s.  1  d.,  5  s.  0  d. 
Write  0  d. 

5  s.,  £1  Is.,  Id.  14  s.,  £2  12  s.,  £3  10  s.,  £4  9  s.,  £5  1  s.,  £5  16  s. 
Write  16  s. 

The  pounds  are  added  as  before. 

57,     Common  Method  of  adding   Compound  Numbers, 

By  the  method  of  adding  compound  numbers  which  is  com- 
monly given,  the  entire  sum  of  each  column  is  found  before 
reducing  to  higher  denominations.  This  method,  however, 
will,  as  a  general  thing,  be  found  much  less  expeditious  than 
.either  of  the  others. 

It  is  illustrated  in  the  following  solution  of  the  example 
given  in  the  last  article. 

Explanation.  —  By  adding  the  farthings  column,  we  find  that  its  sura 
is  13  qr.,  which,  as  4  qr.  =  I  d.,  must  equal  as  many  pence  as  there 
are  times  4  in  13,  which  are  three  times,  with  a  remainder  of  1.  There- 
fore, 13  qr.  =  3  d.  1  qr. 

Writing  1  as  the  farthings  figure  of  the  amount,  we  add  the  3  d.  with 
the  figures  of  the  pence  column :  this  gives  60  d.,  which,  as  12  d.  =  1  s., 
are  equal  to  as  many  shillings  as  there  are  times  12  in  60,  which  are  5 
times.     Therefore,  60  d.  =  5  s. 

Writing  0  as  the  pence  figure  of  the  amount,  we  add  the  5  s.  with  the 
figures  of  the  shillings  column.  This  gives  116  s.,  which,  as  20  s.  =  £1, 
are  equal  to  as  many  pounds  as  there  are  times  20  in  116,  which  are  5 
times,  with  £V  rei])ft}n4?r  of  ]  6,    Therefore.  116  s.  =  £5  1 6  s 


56  ADDITION. 

Writing  16  s.,  we  add  the  £5  with  the  figures  of  the  pounds  column. 
This  gives  £48,  which,  being  the  highest  denomination,  we  write. 

As  all  the  denominations  have  now  been  added,  the  sum  or  amount 
must  be  £48  16  s.  1  qr. 

58.    Examples  for  Practice  in  the  Methods   of  50 5  51, 
and  5G. 

1.  Whtit  is  the  sum  of  £4  17  s.  11  d.  2  qr.  +  £84  13  s.  3  d. 
+  £7  19  s.  8  d.  3  qr.  +  £16  18  s.  9  d.  1  qr.  +  £7  15  s.  1  d. 
+  £18  16s.  lid.  2qr.? 

2.  >yhat  is  the  sum  of  £1386  15s.  6d.  +  £3576  18  s.  10  d. 
+  £463  19  s.  4d.  +  £23  5  s.  8  d.  +  £648  4  s.  6  d.  + 
£100  10  s.  3  d.  ? 

3.  What  is  the  sum  of  40  lb.  7  oz.  5  dwt.  6  gr.  +  9  lb.  8  oz. 
19  dwt.  22  gr.  -\-  2  lb.  11  oz.  19  dwt.  23  gr.  +  7  lb.  8  dwt 
19  gr.  +  11  oz.  6  dwt.  +  3  lb.  1  oz.  15  gr.  +  8  lb.  17  dwt.  + 
31b.  23  gr.  +  18  dwt.  7  gr.  +  9  oz.  15  gr.  +  71b.  3oz. 
13  dwt.  15  gr.? 

4.  What  is  the  sum  of  18  w.  4  da.  21  h.  37  m.  5  sec.  + 

37  w.  5  da.  16  h.  43  m.  57  sec.  +   19  w.  3  da.   14  h.    46  m. 

38  sec.  +  19  w.  6  da.  23  h.  56  m.  27  sec.  +  43  w.  5  da.  2  h. 
17  m.  38  sec.  +  28  w.  1  da.  1  h.  5  m.  7  sec.  ? 

5.  What  is  the  sum  of  47  gal.  3  qt.  1  pt.  2  gi.  +  37  gal. 
1  qt.  1  pt.  1  gi.  +  85  gal.  2  qt.  2  gi.  +  25  gal.  2  qt.  1  pt.  3  gi. 
+  54  gal.  2  qt.  1  pt.  3  gi.  +  18  gal.  2  qt.  1  pt.  2  gi.  +  37  gal. 
3qt.  Opt.   Igi.  +   19  gal.   3  qt.   1  pt.   2gi.  +  43 gal.   0  qt, 

0  pt.  3  gi.  ? 

6.  What  is  the  sum  of  15  yd.  2qr.  2na.  +  18  yd.  3  qr. 

1  na.  +  27  yd.  3  qr.  3  na.  +  42  yd.  1  qr.  +  87  yd.  3  na.  + 
3  qr.  3  na.  +  26  yd.  1  qr.  1  na.  +  57  yd.  3  qr.  2  na-  +  42  yd. 
0  qr.  0  na.  +  64  yd.  3  qr.  3  na.  ? 

7.  What  is  the  sum  of  18  lb  6§  53  2  9  5  gr.  +  7  lb  85 
73  19  18gr.  +  41b  11  §  43  2  9  13  gr.  +  251b  95 
19  4gr.  +  II5  19  +  21b  55  63  09  16gr.  +  51b 
115  45  19  14gr.? 

8.  What  is  the  sum  of  2  m.  7  fur.  28  i-d.  4  yd.  1  ft.  3  in.  + 
6  m.  5  fur.  19  rd.  2  yd.  2  ft.  1 1  in.  +  25  m.  4  fur.  37  rd.  5  yd. 


ADDITION.  ,  57 

Sin.  +  94  ra.  1  fur.  24  rd.  4  yd.  2  ft.  Sin.  +  14  m.  6  fur. 
23  rd.  2  yd.  0  ft.  7  in.  -f  23  m.  5  fur.  37  rd.  4  yd.  1  ft.  10  in. 
-f  57  m.  0  fur.  33  rd.  5  yd.  1  ft.  1  in.  ? 

9.  What  is  the  sum  of  19  m.  5  fur.  37  rd.  2  yd.  2  ft.  2  in. 
-pl6m.  4  fur.  18  rd.  5  yd.  1  ft.  7  in.  +  37  m.  15  rd.  2  yd. 
2  ft.  Sin.  +  17  rd.  5  yd.  7  in.  +  3  m.  7  fur.  18  rd.  4  yd. 
2  ft.  9  in.  +  46  m.  3  fur.  13  rd.  2  yd.  1  ft.  9  in.  +  33  m.  4  fur. 
27  rd.  5  yd.  1  ft.  4  in.  +  19  m.  0  fur.  34  rd.  3  yd.  0  ft.  10  in.  ? 

10.  What  is  the  sum  of  14  A.  3  R.  28  sq.  rd.  27  sq.  yd. 
8  sq.  ft.  12  sq.  in.  +  27  A.  2  R.  31  sq.  rd.  17  sq.  yd.  5  sq.  ft. 
137sq.in.  +  35  A.  1  R.  31  sq.  rd.  18  sq.  yd.  5  sq.  ft.  IIG 
»q.  in.  -|-  21  A.  26  sq.  rd.  25  sq.  yd.  5  sq.  ft.  107  sq.  in.  -|- 
43  A.  2  R.  14  sq.  rd.  19  sq.  yd.  -f  1  R-  15  sq.  rd.  37  sq.  in.  ? 

11.  I  bought  some  flour  for  $6.75  ;  some  cloth  for  $17.25  ; 
a  hat  for  $3.37  ;  a  coat  for  $19.42  ;  a  vest  for  $3.87  ;  some 
calico  for  $3.25  ;  some  flannel  for  $4.93  ;  some  silk  for  $23.99  ; 
a  pair  of  boots  for  $5.33  ;  an  overcoat  for  $22.75  ;  a  shawl 
for  $6.68  ;  a  pair  of  gloves  for  $1.46 ;  an  umbrella  for  $1.37; 
and  a  pair  of  overshoes  for  $1.17.  What  was  the  amount  of 
my  purchase  ? 

12.  A  trader  sold  17  cases  of  broadcloth  ;  the  first  case 
contained  317  yards,  the  second  296,  the  third  319,  the  fourth 
339,  the  fifth  259,  the  sixth  347,  the  seventh  329,  the  eighth 
286,  the  ninth  321,  the  tenth  294,  the  eleventh  337,  the 
twelfth  248,  the  thirteenth  324,  the  fourteenth  346,  the  fifteenth 
299,  the  sixteenth  338,  and  the  seventeenth  207.  How  many 
yards  were  there  in  all  ? 

13.  He  received  $984.36  for  the  first  case,  $849.23  for  the 
second,  $1097.28  for  the  third,  $1342.94  for  the  fourth, 
$836.28  for  the  fifth,  $1297.89  for  the  sixth,  $1048.30  for 
the  seventh,  $857.82  for  the  eighth,  $1004.28  for  the  ninth, 
$976.87  for  the  tenth,  $1248.67  for  the  eleventh,  $827.61  for 
the  twelfth,  $1176.04  for  the  thirteenth,  $1327.98  for  the 
fourteenth,  $876.48  for  the  fifteenth,  $1200.36  for  the  six- 
teenth, and  $758.93  for  the  seventeenth.  How  much  did  he 
receive  for  all  ? 

14.  In  the  course  of  the  year  1853,  a  flour  dealer  bought 


«J8  ADDITION. 

649  barrels  of  flour  for  $3798.75  ;  357  barrels  for  $2039.25 ; 
439  barrels  for  $2679.00  ;  987  barrels  for  $6198.42  ;  299 
barrels  for  $1925.37;  1168  barrels  for  $7385.94;  627  bar- 
rels for  $4369.27  ;  1359  barrels  for  $9967.84 ;  538  barrels 
for  $4279.63 ;  275  barrels  for  $2383.50 ;  96  barrels  for 
$816.00;  948  barrels  for  $8472.56;  358  barrels  for  $3615.80; 
796  barrels  for  $8237.29;  and  2962  barrels  for  $25,851.00. 
How  many  barrels  did  he  buy  in  all  ?  How  many  dollars  did 
he  pay  for  the  whole  ? 

15.  He  gained  $324.50  on  the  first  lot;  $178.50  on  the 
second;  $109.75  on  the  third;  $740.25  on  the  fourth;  $29.90 
on  the  fifth;  $584  on  the  sixth;  $600  on  the  seventh;  $1359 
on  tlie  eighth  ;  $470.75  on  the  ninth  ;  $277.75  on  the  tenth  ; 
$89.28  on  the  eleventh;  nothing  on  the  twelfth  and  thirteenth ; 
$398  on  the  fourteenth ;  and  $2154.25  on  the  fifteenth. 
What  was  the  amount  of  his  gains  ? 

50.    Addition  of  several  Columns  at  one   Operation, 

(a.)  Accountants  often  add  two  or  three,  and  sometimes 
four  or  more,  columns  of  figures  at  a  single  operation. 

(6.)   The  following  illustrates  some  of  the  methods  of  doing  it :  — 

67 

85 

94 

28 

69 

343 

Explanation.  —  69  plus  20  =  89,  plus  8  =  97,  plus  90  =  187,  plus  4 
=  191,  plus  80  =  271,  plus  5  =  276,  plus  60  =  336,  plus  7  =  343. 

(c.)  By  adding  tens  first,  and  then  units,  as  before,  and  naming  only 
results,  we  have  G9,  89,  97,  187,  191,  271,  276,  336,  343. 

{ i.)  A  little  practice  will  enable  a  person  to  add  without  separating 
each  number  into  tens  and  units,  thus:  69,  97,  191,  276,  343. 

(e.)  After  the  student  has  become  familiar  with  the 
method  of  adding  by  single  columns,  he  will  find  it  a  very 
valuable  exercise  to  add  as  above  explained.  We  recommend 
that  he  perform,  at  least,  the  first  twenty  examples  under  5S 
by  adding  two  or  moi-e  columns  at  a  time. 


ADDITION.  .                   59 

60.    Leger  Columns. 

A  great  part  of  the  work  of  an  accountant  consists  in  add- 
ing long  leger  columns,  like  the  following.  Let  the  pupil  find 
the  sum  of  the  numbers  in  each,  being  as  careful  to  obtain  a 
correct  result  as  he  would  be  if  he  were  to  receive  or  pay  the 
several  amounts. 

1.                              2.  3. 

8.37              .78  673.SS 

A.33               .A7  5^7. B A 

7.6s                 .53  34S6.B7 

.A8             S.75  S/'^.aB 

.^7            /.SO  B.37 

S.50               A.37  /67.SA 

6./<^                B.S<p  6<^B6.3S 

/O.OO             /3.85  67 A^. 3/ 

A.SB               S.OO  AS63.S7 

8.07                  .6S  75AS.35 

A.37              .S5  s^86.s8 

<^.a8            /.37  37(^.87 

A.S/                ^.83  S.5f 

/3.S6            6.75  6^.80 

/.SO               8.43  AO60.75 

.57            S0.A8  30<^.7/ 

3.08            6.00       '  /SA.87 

A.96               /.OO  85S0.06 

.85               /.50  SA^3.S8 

A.oo            7.6<^  A8.75 


60 


L» 

ADDITION. 

4. 

6. 

6. 

//AS7.B/ 

2/7B.63 

4^7B.3B 

/3.87 

74B.29 

5/37.96 

49.00 

4.37 

2000.00 

674.00 

69.4B 

/697.B/ 

S.75 

S46.53 

52B.63 

94B3.S5 

5B.47 

542B.49 

27.96- 

697.5B 

954.B6 

4.37 

792.43 

2777.78 

/94B.74 

^246..5B 

934.67 

647.S5 

642. /7 

528.39 

34.9B 

42B.00 

776.95 

S97.S6 

/ooo.oo 

82.55 

49.S7 

3B6.74 

^67.73 

34B.54 

/7.^9 

4/27.4B 

9.7B 

4.26 

29B.49 

6.25 

269.73 

5842.76 

4327.69 

49.47 

378.35 

5/4.3B 

5B3.2B 

-       49.27 

693.27 

679.59 

/B9.O/ 

43.96 

2B74.43 

//0'/.48 

279.B4 

B97.6/ 

698.4/ 

57B6.39 

2B54.55 

64.8/ 

2B4.62 

7443.75 

587.65 

75.2B 

4/.2B 

/4.39 

SUBTRACTION.  W 

SECTION   V. 

SUBTRACTION. 

61  •    Definitions  and  Illustrations. 
(a.)    Subtraction    is    the   process    by   which   we 

FIND    THE    DIFFERENCE    OF   TWO    GIVEN   NUMBERS,    OR  THB 
EXCESS    OF    ONE    GIVEN   NUMBER    OVER   ANOTHER. 

(6.)    The  following  are  questions  in  subtraction  :  — 

Joseph  had  34  apples,  and  gave  away  6  of  them.  How  many  did  he 
have  left  ? 

Samuel  had  16  cents  and  George  had  9.  How  many  more  had 
Samuel  than  George  ? 

8  from  16  leaves  how  many  ? 

How  many  are  12  —  8  ? 

(c.)  The  larger  given  number,  or  one  from  which  we  sub- 
tract, is  called  the  Minuend;  the  smaller  given  number,  or 
one  subtracted,  is  called  the  Subtrahend ;  and  the  result  ob- 
tained is  called  the  Difference  or  Remainder. 

Illustration.  —  In  the  first  of  the  above  examples  34  is  the  minuend,  6 
is  the  subtrahend,  and  28  is  the  diflFerence  or  remainder. 

{d.)  The  minuend  and  subtrahend  must  represent  things  of 
the  same  kind,  otherwise  the  subtraction  cannot  be  performed. 

Illustrations.  —  5  apples  from  7  apples  leave  2  apples,  and  5  pears 
from  7  pears  leave  2  pears ;  but  it  would  be  impossible  to  take  5  pears 
from  7  apples,  or  5  apples  from  7  pears.  We  cannot  subtract  5  cents 
from  7  dimes ;  but  if  we  should  exchange  one  of  the  dimes  for  its  value 
in  cents,  we  should  have  6  dimes  and  10  cents,  from  which  if  we  should 
subtract  5  cents,  there  would  be  6  dimes  and  5  cents  left. 

We  cannot  subtract  units  from  tens,  but  we  can  find  how  many  units 
%  given  number  of  tens  is  equal  to,  and  then  subtract  from  that  number 
of  units. 

63*    Method  of  writing  Numbers  and  performing  Probleim 
requiring  no  Reduction. 

(a.)    Although  the  result  is  not  affected  by  the  manner  of 
writing  the  numbers,  it  is  convenient  to  place  those  of  the  same 
6 


62  SUBTRACTION. 

denomination  near  each  other,  or  to  write  units  under  anits, 
tens  under  tens,  &c.,  in  simple  numbers,  and  pounds  under 
pounds,  shillings  under  shillings,  yards  under  yards,  feet  under 
feet,  &c.,  in  compound. 

(5.)    For  the   sake   of  uniformity,  we   usually  place  the»> 
minuend  above  the  subtrahend,  and  the  remainder  beneath; 
beginning  at  the  right  hand  to  subtract. 

(c.)  The  following  examples  will  illustrate  the  application 
of  these  principles :  — 

1.   Whatisthevalueof  4697-— 3265? 

Solution.  —  Beginning  at  the  right  hand,  and  considering  the  denomi 
nations  separately,  we  have  5  units  from  7  units  leave  2  units ;  6  tens 
from  9  tens  leave  3  tens ;  2  hundreds  from  6  hundreds  leave  4  hundreds ; 
3  thousands  from  4  thousands  leave  1  thousand. 

Having  thus  subtracted  the  numbers  in  all  the  denominations,  we 
know  that  the  remainder  must  be  1  thousand,  4  hundred,  3  tens,  and  2 
units,  or  1432. 

The  work  would  be  written  thus :  — 
4697  =  Minuend. 
3265  =  Subtrahend. 


1432  =  Remainder. 

Second  Example,  —  What  is  the  value  of  £17  8  s.  9  d.  — 
£3  4s.  6d.? 

Solution.  —  Beginning  with  the  lowest  denomination,  we  have  6d. 
from  9  d.  =  3  d. ;  4  8.  from  8  s.  =  4  s. ;  £3  from  £17  =  £14. 
The  answer  is,  therefore,  £14  4  s.  3  d. 

Or,  beginning  at  the  left,  £17  —  £3  =  £14 ;  8  s.  —  4  s.  =  4  s. ;  9  d. 
—  6d.  =  3d.    ^ns.  £14  4  8.  3d. 

The  work  would  be  written  thus :  — 
£.     8.     d. 

17     8     9     Minuend. 
8    i^  #    :Bubtrahend. 


:^'' 


14     4     3     Remainder. 

63.    Methods  of  ProoJ, 
(a.)    From  the  nature  of  subtraction,  it  is  evident,  that  if 
the   minuend  were   divided   into   two   paits,  such  that  one 


SUBTRACTION.  63 

should  equal  the  subtrahend,  the  other  would  equal  the  re- 
mainder. 

(b.)  We  have,  therefore,  the  following  methods  of  test- 
ing the  correctness  of  the  work :  — 

First  Method,  —  Add  the  remainder  to  the  subtrahend,  and 
if  the  sum  thus  obtained  is  equal  to  the  minuend,  the  work  is 
probably  connect ;  but  if  it  is  not,  there  is  an  error  in  either 
the  subtraction  or  the  addition,  and  possibly  in  both. 

Second  Method.  —  Subtract  the  remainder  from  the  minu- 
end, and  if  the  result  thus  obtained  equals  the  subtrahend, 
the  work  is  probably  correct. 

WRITTEN    WORK   AND    PROOF    OF    THE   FIRST    EXAMPLE. 

4697     Minuend. 
3265     Subtrahend. 


1432     Remainder. 


4697     Sum  of  Rem.  and  Sub.  =  Minuend. 


3265     Diffl  of  Rem.  and  Min.  =  Subtrahend. 

WRITTEN   WORK  AND   PROOF    OF   SECOND    EXAMPLE. 
£.      s.     d. 
17     8     9    Minuend. 

3     4     6    Subtrahend. 


14     4    3     Remainder. 


17     8     9    Sum  of  the  Sub.  and  Rem.  =  Minuend. 


3     4     6    Difference  of  Rem.  and  Min.  =  Subtrahend. 

64:.    Problems  requiring  no  Reduction, 
What  is  the  value  of  each  of  the  following  ? 

1.  854736  —  721423  ? 

2.  9863764  —  420423  ? 

3.  2948769  —  1432526? 

4.  $5476.92  — $1261.40? 


64  SUBTRACTION. 

5.  736.87  yd. —  415.247(1.? 

6.  698.795bu.  — 243.521  bu.? 

7.  3  lb.  11  oz.  15  dwt.  18  gr.  —  1  lb.  8  oz.  13  dwt.  4gr.? 

8.  6  T.  17  cwt.  2  qr.  26  lb.  13  oz.  11  dr.  —  2  T.  6  cwt. 
15  lb.  8  oz.  3  dr.  ? 

9.  161b8§  53  2  9  18gr.  —  5ft)4§  2  5  1  9  6gr.? 

10.  1757  gal.  3  qt.  1  pt.  3  gi.  —  1323  gal.  1  qt.  2  gi.  ? 

11.  18 rd.  4yd.  2ft.  llin.  —  6  rd.  2yd.  1ft.  5in.? 

65.    Simple  Subtraction,     Method  when  Reductions  are 
necessary. 

When  *as  is  often  the  case,  a  figure  in  the  subtrahend  repre- 
sents a  greater  value  than  the  corresponding  figure  of  the 
minuend,  we  take  one  of  a  higher  denomination  in  the  minu- 
end, reduce  it  to  the  required  denomination,  add  its  value  to 
the  value  of  the  figure  already  expressed,  and  subtract  the 
value  of  the  subtrahend  figure  from  the  sum  thus  obtained. 

First  Example.  —  What  is  the  difference  between  62.7 
and  35.86  ? 

WRITTEN   "WORK. 

Minuend,  changed  in  form. 

Minuend. 

Subtrahend. 

2     6.84       Difference. 

Explanation  of  Process.  —  As  there  are  no  hundredths  expressed  in 
the  minuend,  we  reduce  one  of  the  7  tenths  to  hundredths,  leaving  6 
tenths.  1  tenth  =  10  hundredths,  from  which  subtracting  6  hundredths, 
leaves  a  remainder  of  4  hundredths. 

As  8  tenths  cannot  be  subtracted  from  6  tenths,  we  reduce  one  of  the 
2  units  to  tenths,  leaving  1  unit.  1  unit  =  10  tenths,  which  added  to 
the  6  tenths  left  in  the  tenths'  place  equal  16  tenths;  8  tenths  from  16 
tenths  =  8  tenths. 

As  5  units  cannot  be  taken  from  the  1  unit  left  in  the  units'  place, 
we  reduce  one  of  the  6  tens  to  units,  leaving  5  tens.  1  ten  =  10  units, 
which  added  to  1  unit  equal  11  units;  5  units  from  11  units  =  6  units 
8  tens  from  5  tens  leave  2  tens. 


0 

11 

.  n 

>,1( 

6 

2 

.7 

3 

5 

.8 

6 

SUBTRACTION.  6§ 

The  answer,  then,  is  2  tens,  6  units,  8  tenths,  and  4  nnndredths,  or 
26.84. 

This  may  be  proved  in  the  same  way  that  the  preceding  examples 
were. 

Qicesttons  on  the  above. 

Which  expresses  the  larger  number,  7  tenths,  or  6  tenths  and  10  hun- 
dredths, and  why?  2  units  and  6  tenths,  or  1  unit  and  16  tenths  1  5 
tens  and  1  unit,  or  4  tens  and  1 1  units  1  6  tens,  2  units,  and  7  tenths, 
or  5  tens,  11  units,  16  tenths,  and  10  hundredths? 

How  then  will  the  remainder,  obtained  by  subtracting  35.86  from 
62.7,  compare  with  the  remainder  obtained  by  subtracting  it  from  5  tens, 
11  units,  16  tenths,  and  10  hundredths  ? 

Second  Example.  —  How  many  are  83004  dollars  minus 
24765  dollars  ? 

WRITTEN    "WORK. 

7     12     9     9     14  =  Minuend,  changed  in  form. 
$8       3     0     0       4  =:  Minuend. 
$2       4     7     6       5  =  Subtrahend. 


$5       8     2     3       9  =  Remainder. 

Explanation.  —  As  we  cannot  take  5  dollars  from  4  dollars,  and  as 
there  are  no  tens  or  hundreds  expressed  in  the  minuend,  we  take  1  thou- 
sand from  the  3  thousands,  leaving  2  thousands  ;  1  thousand  =  10  hun- 
dreds, and  taking  I  of  these  hundreds  to  reduce  to  tens,  we  have  9 
hundreds  left.  1  hundred  ^10  tens,  and  taking  1  of  these  tens  to 
reduce  to  units  we  have  9  tens  left.  1  ten  =  10  units,  which  added  to 
the  4  units  in  the  units'  place  =  14  units. 

Now,  by  subtracting,  we  have 

9  units  from  14  units  =  5  units. 

6  tens  from  9  tens  =:  3  tens. 

7  hundreds  from  9  hundreds  =  2  hundreds. 

4  thousands  cannot  be  taken  from  2  thousands,  therefore  we  take  1 
ten-thousand  from  the  8  ten-thousands  in  the  minuend,  leaving  7  ten- 
thousands,  1  ten-thousand  =  10  thousands,  which  added  to  the  2 
thousands  in  the  thousands'  place  =  12  thousands. 

4  thousands  from  12  thousands  =  8  thousands. 

2  ten-thousands  from  7  ten-thousands  =  5  ten-thousands. 

The  remainder  is,  therefore,  58239  dollars. 

Questions  upon  the  above.  —  How  can  it  be  shown  that  3004  is  equal 
to  2  thousands,  9  hundreds,  9  tens,  and  14  units  ?     That  83004  is  equal 
to  7  ten-thousands,  12  thousauas,  9  hundreds,  9  tens,  and  14  units  ? 
6* 


,66  SUBTRACTION. 

OG.     Compound  Subtraction, 

First  Example.  —  Find  the  diflference  between  141b.  6oz. 
5  dwt.  7  gr.  and  8  lb.  7  oz.  18  dwt,  23  gr. 


WRITTEN    WORK. 

13 

17 

34 

41     Minuend  changed  in  form. 

lb. 

oz. 

dwt. 

gr. 

U 

6 

15 

17     Minuend. 

8 

7 

18 

23     Subtrahend. 

5     10     16     18     Remainder. 

Explanation.  —  As  we  cannot  subtract  23  gr.  from  1 7  gr.,  we  take 
1  dwt.  from  the  15  dwt.,  which  reduced  to  grains  and  added  to  the  17  gr. 
equals  41  gr. ;  23  gr.  from  41  gr.  =  18  gr. 

But  as  18  dwt.  cannot  be  taken  from  the  14  dwt.  left  in  the  minuend, 
we  take  1  oz.  from  the  6  oz.  and  reduce  it  to  pennyweights ;  I  oz.  = 
20  dwt.,  which  added  to  the  14  dwt.  equal  34  dwt. ;  18  dwt.  from  34  dwt 
=  6  dwt. 

As  7  oz.  cannot  be  taken  from  the  5  oz.  left  in  the  minuend,  we  take 
1  lb.  from  the  14  lb.  and  reduce  it  to  ounces  ;  1  lb.  =  12  oz.,  which  added 
to  the  5  oz.  equal  17  oz. ;  7  oz.  from  17  oz.  =  10  oz. 

8  lb.  from  13  lb.  =  5  lb. 

The  answer  is,  therefore,  5  lb.  10  oz.  6  dwt.  8  gr.,  which  may  be  proved 
as  before. 

Questions.  —  Which  expresses  the  greater  quantity,  15  dwt.  17  gr.,  or 
14  dwt.  41  gr.,  and  why  1  6  oz.  14  dwt.,  or  5  oz.  34  dwt.  ?  14  lb.  5  oz., 
or  13  lb.  17  oz.?  14  lb.  6  oz.  15  dwt.  17  gr.,  or  13  lb.  17  oz.  34  dwt. 
41  gr.  1 

How  would  the  remainder  obtained  by  subtracting  8  lb.  7  oz.  18  dwt. 
23  gr.  from  14  lb.  6  oz.  15  dwt.  17  gr.  compare  with  that  obtained  by 
subtracting  it  from  13  lb.  17  oz.  34  dwt.  41  gr.  1 

Second  Example,  —  A  farmer  took  8  bu.  3  pk.  5  qt  of 
corn  from  a  bin  containing  17  bushels.  How  many  bushels, 
pecks,  and  quarts  remained  ? 

Reasoning  Process.  —  If  the  bin  contained  18bn.,  and  he  took  out 
8bu.  3pk.  5  qt.,  there  would  remain  the  difference  between  17  bu.  and 
8  bu.  3  pk.  5  qt.  This  shows  that  17  bu.  is  the  minuend,  and  8  bu.  3  pk 
5  qc.  the  subtrahend. 


SUBTRACTION.  67 

16  3     8     Minuend,  changed  in  form, 
bu.    pk.   qt. 

17  0     0     Minuend. 

8     3     5     Subtrahend. 


8     0     3     Remainder. 

Explanation.  —  As  there  are  no  pecks  or  quarts  expressed  in  the 
minuend,  we  take  1  bu.  from  the  17  bu.  and  reduce  it  to  lower  denomina- 
tions. 1  bu.  =  3  pk.  8  qt.  Therefore  17  bu.  =  16  bu.  3  pk.  8  qt.  The 
subtraction  can  now  be  performed  as  before. 

(a.)  In  examples  involving  fractional  denominations,  it  will 
usually  be  more  convenient  to  make  all  the  reductions  and 
changes  in  the  minuend  before  beginning  to  subtract,  as  in  the 
folloviring  :  — 

Third  Example.  —  What  is  the  difference  between  8  rd. 
3  yd.  1  ft.  4  in.  and  2  rd.  4  yd.  2  ft.  5  in.  ? 

WRITTEN    WORK. 

7  8     2  10  =  Minuend,  changed  in  form. 
rd.    yd.    ft.    in. 

8  3     14:=  Minuend. 

2     4     2     5  =  Subtrahend. 


5     4     0     5=:  Remainder. 

Explanation.  —  Since  there  are  more  yards,  feet,  and  inches  expressed 
in  the  subtrahend  than  in  the  minuend,  we  will  take  1  rd.  from  the  8  rd. 
and  reduce  it  to  lower  denominations.  1  rd.  =  5^  yd.  =  5  yd.  1  ft.  6  in., 
which  added  to  the  3  yd.  1  ft.  4  in.  equal  8  yd.  2  ft.  10  in.  Therefore 
8  rd.  3  yd.  1  ft.  4  in.  —  2  rd.  4  yd.  2  ft.  5  in.  =  7  rd.  8  yd.  2  ft.  10  in.  — 
2  rd.  4  yd.  2  ft.  5  in. 

Note.  —  Had  not  the  1  rod  been  reduced  to  yards,  and  the  ^  yard  to 
feet  and  inches,  before  commencing  the  subtraction,  the  answer  would 
have  taken  the  form  of  5  rd.  3i  yd.  1  ft.  11  in.,  from  which,  by  reducing 
the  i  yard  to  feet  and  inches,  we  should  get  5rd.  3  yd.  2  ft.  17  in.  =  5  rd. 
i  yd.  0  ft.  5  in.  =  the  answer  obtained  directly  by  first  method. 


G7.    Problems  for  Solution 
What  is  the  value  of  — 

1.  3743  —  2554?. 

2.  839.74  —  213.78? 


3.  37.9623  —  21.978? 

4.  300.67  —  25.38  ? 


68  JjUIlTRACTION. 


9.  40006  —  14308  ? 

10.  294.6  —  87.942? 

11.  320.06  —  5.947? 

12.  824.57  —  347.28  ? 


5.  .7235  —  .0649  I 

6.  .00632  —  .00274  ? 

7.  87432  —  52841  ? 

8.  92846  —  24547  ? 
What  is  the  value  of  — 

13.     £17  6  s.  8  d.  1  qr.  —  £13  17  s.  3  d.  2  qr.  ? 
^     14.     25  cwt.  1  qr.  13  lb.  10  oz.  7  dr.  —  13  cwt.  2  qr.  15  lb 
3oz.  11  dr.? 

15.  8  lb.  4  oz.  17  dwt,  13  gr.  —  2  lb.  8  oz.  19  dwt.  20  gr.  ? 

16.  14  lb  9  §  4  3  2  9  12  gr.  —  4  lb  10  3  4  3  2  9 
16  gr.? 

17.  4147  bu.  1  pk.  4  qt.  —  2878  bu.  2  pk.  7  qt.  1  pt.  ? 

18.  49  w.  3  da.  19  h.  13  m.  45  sec.  —  18  w.  1  da.  22  h. 
40m.  53 sec?  ^ 

19.  £487  6  s.  0  d.  1  qr.  —  £236  11  s.  8  d.  3  qr.  ? 

20.  27°  24'  47"  —  19°  37'  51"  ? 

21.  38  m.  4  fur.  23  rd.  4  yd.  Oft.  3  in.  —  32  m.  5  fur 
28  rd.  5  yd.  1  ft.  5  in.  ? 

22.  54  m.  6  fur.  3  rd.  8  in.  —  48  m.  3  fur.  3  rd.  2  yd. 
2  ft.  10  in.  ? 

23.  3  R.  14  sq.  rd.  7  sq.  yd.  2  sq.  ft.  19  sq.  in.  —  23  sq.  rd. 
11  sq.  yd.  5  sq.  ft.  138  sq.  in.  ? 

24.  15  rd.  5  yd.  2  ft.  11  in.  —  16  rd.  1  ft.  4  in.  ? 

08.     The   Changed  Minuend  not  usually  written, 

(a.)  The  changed  form  of  the  minuend  has  been  written  in 
the  preceding  examples  to  insure  tliat  the  nature  of  the  reduc- 
tions and  changes  shall  be  understood  by  the  pupil.  It  is  not, 
however,  customary  to  write  it.  The  full  explanation  is  the 
same  whether  it  is  written  or  omitted ;  but  when  it  is  omitted, 
and  every  step  of  the  process  is  understood  and  mastered,  ab- 
breviated explanations  like  the  following  may  be  adopted  :  — 

WRITTEN   WORK   OF    FIRST    EXAMPLE    UNDER   6«S. 

62.7    =  Minuend. 
35.86  =  Subtrahend. 


26.84  =  Remainder. 


SUBTRACTION.  69 

First  Abbreviated  Explanation.  —  6  hundredths  cannot  be  taken  from 
0  hundredths,  but  1  tenth  =  10  hundredths,  and  6  hundredths  from  10 
hundredths  leave  4  hundredths. 

8  tenths  cannot  be  taken  from  6  tenihs,  but  one  unit  equal  10  tenths, 
and  6  tenths  added  =  16  tenths.     8  tenths  from  16  tenths  =  8  tenths. 

5  units  cannot  be  taken  from  1  unit,  but  1  ten  =  10  units,  tnd  1  unit 
added  =  11  units.     5  units  from  11  units  leave  6  units. 

3  tens  from  5  tens  leave  2  tens. 

Hence  the  answer  is  26.84. 

Second  Abbreviated  Explanation.  —  6  hundredths  from  10  hundredths 
leave  4  hundredths  5  8  tenths  from  16  tenths  leave  8  tenths ;  5  units  from 
11  units  leave  6  units ;  3  tens  from  5  tens  leave  2  tens. 

This  gives  26.84  for  an  answer,  as  before. 

(h.)  All  explanations  should  finally  be  dropped,  and  only 
results  named,  thus  :  —  4  hundredths,  8  tenths,  6  units,  2  teiis ; 
giving  for  the  answer  26.84,  as  before. 

WRITTEN   WORK  OP   FIRST    EXAMPLE    UNDER   66. 

lb.       oz.  dwt.    gr. 

14       6  15     17  =  Minuend. 

8       7  18     23  =  Subtrahend. 


5     10     16     18  =  Remainder. 

First  Abbreviated  Explanation.  —  23  gr.  cannot  be  subtracted  from 
17  gr. ;  but  1  dwt.  =  24  gr.,  and  17  gr.  added  are  41  gr.  23  gr.  from  41  gr. 
:=18gr. 

18  dwt.  cannot  be  taken  from  14  dwt. ;  but  1  oz.  =  20  dwt.,  and 
14  dwt.  added  =  34  dwt.     1 8  dwt.  from  34  dwt.  =  1 6  dwt. 

7  oz.  cannot  be  taken  from  5  oz. ;  but  1  lb.  =  12  oz.,  and  5  oz.  added 
are  17  oz.    7  oz.  from  17  oz.  =  10  oz. 

Hence  the  remainder  is  5  lb.  10  oz.  16  dwt.  18  gr. 

60.    Subtrahend  Figure  mag  be  increased  instead  of 
diminishing  Minuend. 

It  is  obvious  that  the  result  would  be  the  same,  if,  instead  of  consid- 
ering the  minuend  figure  of  a  denomination  from  which  a  reduction  has 
been  made  to  be  one  less,  we  should  consider  the  corresponding  subtra- 
hend figure  to  be  one  greater.  In  the  former  case,  we  subtract  1  (on 
account  of  the  reduced  unit)  before  subtracting  the  subtrahend  figure 
while  in  the  latter  we  add  1  to  the  subtrahend  figure,  and  subtract  both 
together. 


70  SUBTRACTION. 

Thns,  in  subtracting  the  tenths  of  the  first  example,  we  may  subtract 
1  tenth  from  the  7  tenths  before  subtracting  the  8  tenths,  or  we  may  add 
the  1  tenth  to  the  8  tenths,  and  subtract  both  at  once.  Many  always 
subtract  by  the  last  method.  One  method  is  as  convenient  as  the  other, 
but  the  one  to  which  we  are  most  accustomed  will  seem  the  easiest. 


70.    Another,  and  usually  shorter^  Method  of  subtracting 

Compound  Numbers. 

(a.)  When  in  compound  subtraction  reductions  are  neces- 
sary, and  the  changed  minuend  is  not  written,  one  of  the 
following  methods  is  often,  if  not  usually,  easier  than  that 
hitherto  taken. 

First.  Subtract  from  the  value  of  the  reduced  unit,  and  add 
the  remainder  to  the  minuend  figure ;  or, 

Second.  Subtract  as  many  as  possible  from  the  written 
minuend  figure,  and  the  rest  from  the  value  of  the  reduced 
unit. 

(b.)  Applying  the  first  of  the  above  methods  to  the  example 
just  considered  gives  the  following  work  :  — 

Subtracting  17  of  the  23  gr.  from  the  17gr.  leaves  6gr.  to  be  taken 
from  24  gr.,  (the  value  of  the  reduced  unit.)     6  gr.  from  24  gr.  =  18  gr. 

Subtracting  14  of  the  18dwt.  from  the  14dwt.  leaves  4dwt.  to  be 
subtracted  from  20  dwt.,  (the  value  of  the  reduced  unit.)  4  dwt.  from 
20dwt.  =  16  dwt. 

Subtracting  5  of  the  7  oz.  from  the  5  oz.  leaves  2  oz.  to  be  taken  from 
12  oz.,  (the  value  of  the  reduced  unit,)    2  oz.  from  12  oz.  =  10  oz. 

8  lb.  from  13  lb.  leave  5  lb. 

(c.)  In  practice,  the  above  method  may  be  abbreviated 
thus :  — 

Subtracting  17  of  the  23  gr.  leaves  6gr.,  and  6gr.  from  1  dwt,  or 
24gr.,  =  18gr. 

Subtracting  14  of  the  18  dwt.  leaves  4  dwt,  and  4  dwt  from  1  oz.,  or 
20  dwt.,  =  16  dwt 

Subtracting  5  of  the  7ob.  leaves  2  oz.,  and  2  oz.  from  1  lb.,  or  12  oz., 
■=  10  oz. 

81b.  from  131b.  =  5  lb. 

(«?.)    Abbreviating  still  more,  we  have  — 

17  from  23  =  6,  and  6  gr.  from  1  dwt,  or  24  gr.,  >=  18  gr. 


SUBTRACTION.  71 

14  from  18  =  4,  and  4  dwt.  from  1  oz.,  or  20  dwt,  =  16  dwt. 

5  from  7  =  2,  and  2  oz.  from  1  lb.,  or  12  oz.,  =  10  oz. 

3  lb.  from  13  lb.  ==  6  lb.,  giving,  as  before,  5  lb.  10  oz.  16  dwt.  18  gr. 

(e.)    Adopting  the  second  method,  we  have  — 

23  gr.  from  1  dwt.,  or  24  gr.,  =  1  gr.,  which  added  to  the  17  gr.  gives 
18  gr. 

18  dwt.  from  1  oz.,  or  20  dwt.,  =:  2  dwt.,  which  added  to  the  14  dwt. 
gives  16  dwt. 

7oz.  from  lib.,  or  12oz.,  =  5oz.,  which  added  to  the  5  oz.  giyes 
10  oz. 

8  lb.  from  13  lb.  =  5  lb.,  giving,  as  before,  5  lb.  10  oz.  16  dwt.  18  gr. 

iq^OTB.  —  Practice  will  make  the  student  so  familiar  with  all  these 
methods,  that  he  will  be  able  to  see  at  once  which  is  best  adapted  to  the 
case  he  is  considering. 

Tl  •    Problems  for  Solution 

What  is  the  value  of — 

1.  3743  —  2917? 

2.  94276  —  46324? 

3.  867004  —  328527  ? 

4.  8674.5  —  2594.326  ? 
6.     45842.7  —  6243.984? 


7.  93.46  —  2.78457? 

8.  6.0004  —  4.000563? 

9.  100000000  —  7  ? 

10.  7000000  —  .000007  ? 

11.  847.96  —  47.96823? 


6.     7564.001  —  756.4002?]   12.     487.6307  —  48.76307? 

13.  A  land  company  bought  8479  acres  of  wild  land,  and 
sold  3896  acres  of  it.     How  many  did  they  have  left? 

Reasoning  Process.  —  If  they  bought  8479  acres,  and  sold  3896  acres 
of  it,  they  must  have  left  the  difference  between  8479  acres  and  3896 
acres,  to  find  which  we  subtract  3896  from  8479. 

14.  A  ship  is  valued  at  $27648,  and  its  cargo  at  $49325. 
How  much  more  is  the  cargo  worth  than  the  ship  ? 

Reasoning  Process.  —  If  the  ship  is  worth  $27648,  and  the  cargo  ia 
worth  $49325,  the  cargo  must  be  worth  as  many  dollars  more  than  the 
ship  as  there  are  in  the  difference  between  $49325  and  $27648,  to  find 
which  the  latter  must  be  subtracted  from  the  former. 

15.  Census  returns  show  that  the  United  States  contained 
392982f  inhabitants  in  the  year  1790;  5305941  in  1800; 
7239814  in  1810;  9638191   in  1820;  12866020  in  1830; 


72  SUBTRACTION 

17069453  in  1840;   and  23263488   in   1850.     How  many 
more  did  they  contain  in  1800  than  in  1790  ? 

16.  How  manj  more  in  1810  than  in  1800? 

17.  How  many  more  in  1820  than  in  1810  ? 

18.  How  many  more  in  1830  than  in  1820  ? 

19.  How  many  more  in  1840  than  in  1830  ? 

20.  How  many  more  in  1850  than  in  1840  ? 

21.  How  many  more  in  1850  than  in  1790  ? 

22.  How  many  more  in  1850  than  in  1800  ? 

23.  How  many  more  in  1820  than  in  1790  ? 
What  is  the  value  of — 

24.  27  bu.  2  pk.  2  qt.  —  18  bn.  3  pk.  3  qt.  ? 

25.  83  yd.  2  qr.  1  na-  —  47  yd.  3  qr.  2  na.  ? 

26.  8cwt.   2qr.    151b.  7oz.  3  dr.  —  2  cwt.  3  qr.  241b. 
8  oz.  9  dr.  ? 

27.  37  gal.  2  qt.  1  pt.  2  gi.  —  12  gal.  3  qt.  1  pt.  3  gi. 

28.  37  lb.  2  dwt.  —  4  lb.  7  oz.  5  dwt.  13  gr.  ? 

29.  24  !b3§  —51b  Hi  45  2  9  14gr.? 

30.  83  yd.  —  2  qr.  3  na.  ? 

31.  £64  — £28  14  s.  7d.  2qr.? 

32.  187  T.  3  qr.  13  lb.  —  67  T.  17  cwt  1  qr..  18  lb.  5  oz. 
13  dr.? 

33.  27  sq.  rd.  5  sq.  ft.  17sq.  in.  —  26sq.  rd.   30  sq.  yd. 
7  sq.  ft.  53  sq.  in.  ? 

34.  6fur.  8rd.  3yd.  1ft.  1  in.  —  1  fur.  8  rd.  4yd.  2  ft. 
11  in.? 

35.  18  yd.  Ina.  —  14  yd.  2qr.  3na.? 

36.  6  m.  1  ft.  —  5  m.  7  fur.  39  rd.  5  yd.  1  ft.  2  in.  ? 

37.  18  m.  —  17  m.  7  fur.  39  rd.  5  yd.  1  ft.  5  in.  ? 

38.  231  A.  19  sq.  rd.  —  197  A.  3  R.  27  sq.  rd.  15  sq.  yd. 
5  sq.  ft.  97  sq.  in.  ? 

39.  127°  18' 14" —  113°  47' 25"? 

40.  307  T.  8  cwt.  2  qr.  23  lb.  8oz.  12  dr.  —  213  T.  15  cwt 
231b.  lloz.  6  dr.? 

41.  527  yd.  1  qr.  2  na.  1  in.  —  431  yd.  2  qr.  3  na.  2  in.  ? 

42.  63  m.  —  27  m.  7  fur.  39  rd.  5  yd.  2  ft.  3  in.  ? 


SUBTRACTION. 


7S 


48.  Bought  7  T.  14cwt.  1  qr.  19  lb.  of  hay,  from  which  T 
sold  3  T.  7  cwt.  3  qr.  26  lb.     How  much  had  I  left  ? 

44.  A  goldsmith  bought  7  lb.  7  oz.  of  gold.  How  much 
will  he  have  left  after  manufacturing  and  selling  3  lb.  5  gr. 
of  it? 

45.  A  trader  sold  9  yd.  3  qr.  2  na.  of  cloth  from  a  piece 
containing  27  yd.  Iqr.  Ina.  How  much  was  left  in  tlio 
piece  ? 

46.  A  man  set  on  foot  to  travel  from  Boston  to  Springfield, 
the  distance  being  98  miles.  The  first  day  he  travelled  28  m. 
7  fur.  19  rd.,  the  second  24  m.  6  fur.  28  rd.,  the  third  29  m. 
4  fur.  36  rd.  How  far  was  he  from  Springfield  at  the  end 
of  the  third  day  ? 

47.  A  man  undertook  to  dig  a  ditch  for  a  certain  price  per 
rod.  On  completing  it  he  demanded  payment  for  a  ditch 
37  rd.  0  ft.  3  in.  long.  His  employer,  doubting  his  honesty, 
measured  it,  and  found  it  to  be  but  36  rd.  5  yd.  1  ft.  9  in.  long. 
A  dispute  arising  between  them,  they  called  in  Mr.  Jeiiks  to 
settle  it,  agreeing  to  abide  by  his  decision.  He  measured  the 
ditch,  and  found  its  length  to  be  36  rd.  16  ft.  9  in.  What  was 
the  difference  in  their  measurements  ? 


TS,    Subtraction  from  Left  to  Right. 

We  can  begin  at  the  left  to  subtract  as  well  as  at  the  right,  if  we  am 
only  careful  to  reserve  one  for  reduction  from  each  denomination  in  the 
minuend  when  it  is  required  by  the  lower  denominations.  This  reduc- 
tion will  be  necessary  when  the  figures  in  the  subtrahend  at  the  riglit  of 
the  denomination  considered  are  greater  than  the  corresponding  ones  of 
the  minuend. 

Example.  —  How  many  are  508.935  —  249.748  ? 
508.935  =  Minuend. 
249.748  =  Subtrahend. 


259.187  =  Remainder. 

Explanation.  —  Eeservmg  1  hundred  from  the  5  hundreds,  we  have, 
2  hundreds  from  4  hundreds  =  2  hundreds.    Reducing  the  1  hundred 
reserved  to  tens,  and  reserving  1  ten  for  further  reduction,  we  have,  4 
7 


74  ^  SUBTRACTION. 

tens  from  9  tens  =  5  tens.  Reducing  the  1  ten  to  units,  and  adding  8 
mits,  we  have,  9  units  from  18  units  =  9  units.  Reserving  1  tenth,  we 
have  7  tenths  from  8  tenths  =  1  tenth.  Reducing  the  1  tenth  reserved  to 
hundredths,  and  adding  2  hundredths,  (1  hundredth  being  reserved,)  we 
have,  4  hundredths  from  12  hundredths  =  8  hundredths.  Reducing  1 
hundredth  to  thousandths,  and  adding  the  5  thousandths,  we  have,  8 
thousandths  from  15  thousandths  =  7  thousandths.  The  answer  is, 
therefore,  259.187. 

As  soon  as  the  process  is  sufficiently  well  understood  to  justify  it,  the 
explanations  should  be  omitted,  and  the  results  only  named. 

Thus,  in  the  example  explained  above,  the  pupil  should  say,  2  hun- 
dreds, 5  tens,  9  units,  1  tenth,  8  hundredths,  7  thousandths.  The  answer 
is,  therefore,  259,187. 

It  is  also  a  good  mental  exercise  to  read  the  results  at  once  by  in- 
specting minuend  and  subtrahend. 

Having  the  numbers  to  be  subtracted  written  thus,  — 
8274.01  =  Minuend, 
2357.23  =  Subtrahend, 
perform  the  subtractions  mentally,  beginning  at  the  left,  and  read  at 
once  5916.78.    Practice  will  make  this  very  easy. 

Perform  the  following  subtractions  by  beginning  at  the  left :  ■— 


1. 

89704  —  29821. 

6. 

521.732  —  23.547. 

2. 

85.1  —  22.563. 

7. 

42.736  —  5.749. 

3. 

$56  —  $8.73. 

8. 

$65.28  —  $47. 

4. 

800635  —  32070. 

9. 

.4103  —  .00627. 

5. 

472968  —  381489. 

10. 

678432  —  189146. 

73.    Suhtr action  of  several  Numbers  at  once. 

A  good  method  of  proceeding  when  several  numbers  are  to 
be  subtracted  is,  to  subtract  the  sum  of  each  column  of  the 
subtrahend  from  the  appropriate  part  of  the  minuend,  redu- 
cing and  changing  denominations,  as  before  explained. 

How  many  are  862  —  28  —  59  —  38  —  56  ? 

WRITTEN    WORK. 

862     Minuend. 
28 


59 
38 
b^  J 


Subtrahends. 


681     Remainder. 


SUBTRACTION.  .  75 

Explanation.  —  Adding  the  units  of  the  subtrahends,  we  have,  6,  14, 
23,  31  units,  which  cannot  be  taken  from  2  units.  To  obtain  as  many 
as  31  units,  we  must  take  3  tens  from  the  6  tens.  3  tens  =  30  units, 
which  added  to  the  2  units  equals  32  units  ;  31  from  32  leaves  1. 

Now,  it  makes  no  difference  whether  we  take  3  tens  (on  account  of 
those  we  reduced  to  units)  from  the  6  tens,  and  afterwards  take  the  tens 
in  the  tens  column  of  the  subtrahend,  or  whether  we  take  all  together. 
Adopting  the  latter  method,  and  adding  the  tens  column,  we  have  3,  8, 
11,  16,  18  tens,  which  cannot  be  taken  from  6  tens,  but  reducing  2  hun- 
dreds to  tens,  and  adding  the  6  tens,  we  have  26  tens,  from  which  18 
tens  being  taken,  there  will  remain  8  tens, 

2  hundreds  from  8  hundreds  =  6  hundreds.  The  answer  is,  there- 
fore, 681. 

When  the  above  ig  well  understood,  omit  in  practice  a  part 
of  the  explanation,  as  follows  :  — 

6,  14,  23,  31  from  32  leaves  1  unit;  3,  8,  11,  16,  18  tens  from  26  tens 
=  8  tens  ;  2  hundreds  from  8  hundreds  =  6  hundreds. 

The  following  form  may  also  be  taken  :  — 
6,  14,  23,  31,  and  1  are  32  units ;  write  1  in  the  units'  place ; 
3,  8,  11,  16,  18,  and  8  are  26  tens ;  write  8  in  the  tens'  place; 
2  and  6  are  8  hundreds  ;  write  6  in  the  hundreds'  place. 

Perform  the  operations  indicated  in  the  following  examples 
by  the  method  explained  above  :  — 

1.  87642  —  273  —  4827  —  285  —  437  —  869  —  245. 

2.  98402  —  2701  —  2596  —  1874  —  987  —  1283  — 
5876. 

3.  276.385  —  31.278  —  13.691  — .  12.586  —  57.84  — 
32.798  —  7.302. 

4.  287000  —  328.7  —  221.3  —  344.37  —  2.851  — 
117.06  —  576.823. 

5.  283.587  —  1.27  —  .328  —  9.063  —  57.063  —  .00876 
_  70.07  —  .826. 

6.  Subtract  .87  +  4.73  +  826  +  42.71  -f  9.854  +  3.27 
from  9012031. 

7.  Subtract  8837  +  1429  +  6372  -f  8406  +  9785  + 
4203  from  9120301. 

8.  Subtract.  8375.94  +  276.483  +  5427.98  +  386.421  + 
279.43  +  81  679  +  .4237  +  4598.7  from  846271.3. 


76  MULTIPLICATION. 

9.  Mr.  Ewell  has  in  his  possession  $9478.63,  but  he  owes 
$143.27  to  Mr.  Webster,  $549.71  to  Mr.  May,  $581,375  to 
Mr.  Kingsbury,  and  $378,875  to  Mr.  Bryant.  How  much 
will  he  have  left  after  paying  his  debts  ? 

10.  A  man  travelled  8725.67  miles  in  the  following  con- 
veyances, viz. :  1285.89  miles  in  railroad  cars,  876.81  miles 
in  a  canal  boat,  587.86  miles  in  a  stage  coach,  725.18  miles  on 
horseback,  647.25  miles  on  foot,  3147.82  miles  in  a  steam- 
boat, and  the  rest  in  a  ship.  How  many  miles  did  he  travel 
in  a  ship  ? 

11.  Messrs.  Howes  and  Baker  bought  27147  bushels  of 
Indian  corn.  After  selling,  at  private  sale,  1438  bushels  to 
one  man,  2627  to  another,  3781  to  another,  and  864  to  an- 
other, they  sold  the  rest  at  auction.  How  many  bushels  did 
they  sell  at  auction?  They  received  $719  for  the  first  lot, 
$1313.50  for  the  second,  $1890.50  for  the  third,  $432  for  the 
fourth,  and  enough  to  make  up  $13573.50  for  what  they  sold 
at  auction.  How  much  did  they  receive  for  that  which  they 
sold  at  auction  ? 


SECTION  VL 

MULTIPLICATION. 

74:.    Definitions  and  Illustrations, 
(a.)    Multiplication    is  a  process    by  which   tte 

ASCERTAIN  HOW  MUCH  ANT  GIVEN  NUMBER  WILL  AMOUNT 
TO,  IF  TAKEN  AS  MANY  TIMES  AS  THERE  ARE  UNITS  IN 
SOME    OTHER    GIVEN    NUMBER. 

(h.)    The  following  are  questions  in  multiplication  :  — 

How  many  are  7  times  6  ? 

What  is  t\k9  valua  of  9  multiplied  by  6  1 


MULTIPLICATION.  77 

What  is  the  value  of  7  X  5  ? 

How  much  will  8  books  cost  at  3  dollars  apiece  1 

(c.)  The  number  supposed  to  he  taken  is  called  the  multi" 
pUcand^  the  number  showing  how  many  times  the  multiplicand 
is  supposed  to  be  taken  is  called  the  multiplier^  and  the  result 
is  called  the  product. 

(d.)  The  multiplier  and  multiplicand  are  called  factors  of 
the  product. 

(e.)    The  product  is  said  to  be  a  multiple  of  its  factors. 

Illustrations.  —  In  the  first  of  the  above  examples,  6  is  the  multipli- 
cand, 7  is  the  multiplier,  and  the  answer,  42,  is  the  product.  7  and  6  are 
factoi's  of  42,  and  42  is  a  multiple  of  7  and  6, 

In  the  second  example,  9  is  the  multiplicand,  6  is  the  multiplier,  and 
the  answer,  54,  is  the  product.  9  and  6  are  factors  of  54,  and  54  is  a 
multiple  of  9  and  6. 

(/)  The  last  example  would  be  solved  thus  :  — 

If  1  book  costs  3  dollars,  8  books  will  cost  8  times  3  dollars,  which 
are  24  dollars. 

Here  3  is  the  multiplicand,  8  is  the  multiplier,  and  24  is  the  product. 
8  and  3  are  factors  of  24,  and  24  is  a  multiple  of  3  and  8. 

(g.)  In  performing  the  operation,  the  multiplier  must 
always  be  regarded  as  an  abstract  number. 

Illustration.  —  A  number  can  be  taken  3  times,  5  times,  or  8  times, 
but  it  would  be  absurd  to  speak  of  taking  it  3  bushels  times,  5  houses 
times,  or  8  books  tirnes. 

(Ji.)  The  product  must  be  of  the  same  denomination  as 
the  multiplicand. 

Illustration.  —  7  times  8  bushels  =  56  bushels ;  9  times  4  books  = 
36  books  ;  7  times  5  tenths  =  35  tenths,  &c. 

Note.  —  It  must  be  observed,  that  there  is  an  apparent  exception  to 
the  last  statement,  {h.)  when  the  multiplier  is  a  fraction,  for  .6  times  .04 
»=  .024  ;  .02  times  .0003  =  .000006,  &c.  This  will  be  explained  in  the 
ijection  on  fractions.     (See  page  173,  Note.) 

(z.)  To  examine  the  nature  of  the  operation  on  the  numbers,  let  us 
suppose  that  a  person  ignorant  of  all  numerical  processes,  except  that 
of  countincj.  should  be  called  upon  to  solve  the  last  question. 

7  * 


78  MULTIPLICATION. 

(j.)  If  he  had  a  quantity  of  dollars,  he  might  lay  3  in  one  place,  S 
more  in  another,  3  more  in  another,  and  so  go  on  laying  3  in  a  place 
till  he  should  have  8  piles  of  3  dollars  each.  Since  the  dollars  in  each 
pile  would  buy  1  book,  the  dollars  in  all  would  buy  8  books ;  he  might 
then,  by  counting  the  dollars  in  the  8  piles,  find  how  much  the  books 
would  cost. 

(k.)  If  he  should  have  no  dollars,  he  might  still  determine  the  result 
in  a  similar  way,  by  using  pebbles,  sticks,  marks,  or  any  thing  else  of  a 
like  cliaracter.  After  learning  how  to  add,  he  might  obtain  the  result 
by  adding  8  threes  together. 

(/.)  If,  aftCK  having  obtained  the  result  in  some  way  similar  to  the 
above,  he  should  remember  it,  he  would  ever  after  be  able,  loithout  count- 
inrj  or  adding,  to  give  the  answer  to  any  question  requiring  the  amount 
of  3  taken  8  times. 

{m.)  If  he  should  learn  in  a  similar  manner  the  several  amounts  of 
10  and  each  number  below  10,  taken  as  many  times  as  there  are  units  in 
each  successive  number  from  1  to  10,  he  would  learn  the  common  mul- 
tiplication table  as  far  as  ten.  If  he  should  now  learn  how  to  apply 
this  knowledge  to  the  decimal  system  of  numbers,  he  would  be  master 
of  the  process  of.  multiplication. 

Note.  —  The  very  common  definition.  "  Multiplication  is  a  short 
method  of  addition,"  is  not  a  good  one,  any  more  than  would  be,  "  Mul- 
tiplication is  a  short  method  of  counting;"  for  while  it  is  true  that  the 
results  obtained  by  multiplication  might  be  obtained  by  addition,  it  is 
equally  true  tnat  they  might  be  obtained  by  counting.  It  is  true  that 
multiplication  has  a  dependence  both  on  addition  and  counting,  but  it  is 
equally  true  that  it  is  as  distinct  from  them  as  they  are  from  each  other, 
and  that  when  we  multiply  we  neither  add  nor  count. 

For  instance,  when  we  find  the  sum  of  75798  -f  24687  +  39764  -f- 
86328  -f-  4395  -|-  283  +  86536,  by  the  method  explained  in  article  50,  we 
add  them ;  but  when  we  merehj  remember^  and  state  that  their  sum  is 
317791,  we  do  not  add  them. 

So  when  we  call  to  ^^mind  that  4  and  4  are  8,  and  4  are  12,  and  4  are 
16,  and  4  are  20,  we  add  5  fours  together ;  but  when  we  merely  remem- 
ber that  5  fours,  or  5  times  4,  are  20,  we  perform  no  addition,  although 
as  a  result,  we  have  in  the  mind  the  sum  of  5  fours.   . 


75.    Product  not  affected  hy  Change  in   Order  of  Factors 

{a,)  In  determining  the  product  of  two  numbers,  it  makes 
no  diiTerence  which  is  regarded  as  the  multiplicand,  provided 
the  other  is  regarded  as  the  multiplier. 


MULTIPLICATION.  79 

Thus  :  6  times  4  =  4  times  6,  or  6  fours  =  4  sixes  =  24. 
Again  :  5  times  3  =  3  times  5,  or  5  threes  =  3  fives  ^15. 

(6.)   The  principle  may  be  proved  true  for  all  numbers,  by  the  follow 
ing  arrangement  of  dots  :  — 


Considering  the  dots  ajs  being  arranged  in  horizontal  rows,  there  are 
3  rows  with  5  dots  in  each  row ;  considering  them  as  being  arranged  in 
vertical  rows,  there  are  5  rows  with  3  dots  in  each  row  ;  and  reckoning 
in  either  way  we  include  all  the  dots. 

(c.)  Now,  if  these  rows  were  extended  in  either  direction,  always 
being  kept  equal  to  each  other,  it  is  evident  that  the  number  of  rows 
reckoned  in  one  direction  would  always  be  equal  to  the  number  of  dots 
that  would  be  in  a  row  were  the  rows  reckoned  in  the  other  direction, 
and  that  all  the  dots  would  be  reckoned  in  both  instances.  The  number 
that  represents  the  multiplicand  when  the  rows  are  reckoned  in  one 
direction,  will  represent  the  multiplier  when  they  are  reckoned  in  the 
other,  while  the  product,  or  number  of  dots,  will  be  unaltered. 

(d.)  Hence,  it  must  always  be  true  that  it  makes  no  difference  with  the 
product  which  of  the  two  factors  is  taken  for  a  multiplier,  provided  the 
other  be  taken  as  the  multiplicand.  It  will  generally  be  most  convenient 
to  consider  the  larger  factor  as  the  multiplicand,  though  not  always  so. 

Note.  —  In  changing  the  order  of  factors,  the  one  taken  for  the 
multiplier  should  always  be  regarded  as  an  abstract  number,  (see  74L,g,) 
while  the  other  should  take  the  denomination  of  the  original  multipli- 
cand. Thus,  4  times  3  apples  =  12  apples  ;  or  changing  the  order  of 
the  factors,  we  should  have  3  times  4  apples  =  12  apples.  In  the  first 
case,  4  is  an  abstract,  and  3  a  concrete,  number;  but  in  the  second,  4  is 
a  concrete,  and  3  an  abstract,  number. 

So  6  times  S8  =:  8  times  $6  ;  4  times  $.09  =  9  times  $.04 ;  5  times 
$4.6./  =  469  times  $.05  ;  &c. 


76.    Simple  MuUipUcation.  —  When   only  one  Factor  is   a 
large  Number. 

When  either  factor  is  a  large  number,  it  will  be  well  to 
consider  its  denominations  separately,  and,  if  we  write  the 
resplts  as  we  obtain  them,  to  begin  with  the  lowest  denomina- 
tion. 

What  will  7  acres  of  land  cost  at  $75.69  per  acre  ? 


80  MULTIPLICATION, 

Reasoning  Process.  —  If  1  acre  costs  $75.69,  7  acres  will  cost  7  times 
$75.69,  which  can  be  found  by  multiplying  it  by  7. 

In  performing  the  requisite  multiplication,  the  numbers  are  usually 
written  in  some  convenient  way,  as  the  following :  — 

$75.69  =  Multiplicand. 
7  =  Multiplier. 

$529.83  =  Product. 

Explanation.  —  Beginning  at  the  right  hand,  or  lowest  denomination 
of  the  multiplicand,  we  have  7  times  9  hundredths  =  63  hundredths,  or 
6  tenths  and  3  hundredths. 

Writing  the  3  in  the  hundredths'  place,  and  reserving  the  6  tenths  to 
add  to  the  product  of  the  tenths  by  7,  we  have  7  times  6  tenths  =  42 
tenths,  and  6  tenths  added,  =  48  tenths,  =  4  units  and  8  tenths. 

"Writing  the  8  tenths,  and  reserving  the  4  units  to  add  to  the  product 
of  the  units  by  7,  we  have  7  times  5  units  =  35  units,  and  4  units  added, 
=  ^9  units,  =  3  tens  and  9  units. 

Writing  the  9  units,  and  reserving  the  3  tens  to  add  to  the  product 
of  the  tens  by  7,  we  have  7  times  7  tens  =  49  tens,  and  3  tens  added,  = 
52  tens,  which,  being  our  last  product,  we  write.  The  result,  then,  is 
529.83. 

Note.  —  As  soon  as  practicable,  the  explanation  should  be  abbre- 
viated, so  as  to  name  only  results.  Thus,  63  hundredths  ;  42,  48  tenths , 
35,  39  units  ;  49,  52  tens.     Ans.  $529.83. 

77.    Multiplication  of  Compound  Numbers. 
What  will  8  casks  of  wine  cost  at  £3  9  s.  7  d.  per  cask  ? 

Reasoning  Process.  —  If  1  cask  costs  £3  9  s.  7  d.,  8  casks  will  cost  8 
times  £3  9  s.  7  d.,  which  can  be  found  by  multiplying  it  by  8. 

WRITTEN    WORK. 
£.     8.     d. 

3       9     7  =  Multiplicand. 
8  =  Multiplier. 


27     16     8  =  Product. 

Explanation.  —  Beginning  with  the  lowest  denomination,  we  have  8 
times  7  d.  =  56  d.,  which,  since  12  d.  =  1  s.,  must  equal  as  many  shil- 
lings as  there  are  times  12  in  56,  which  are  4  times,  with  a  remainder  of 
8.     Hence,  56  d.  =  4  s.  8  d. 

Writing  the  8  d.,  and  reserving  the  4  s.  to  add  to  the  shillings  of  the 


MULTIPLICATION.  81 

next  product,  we  have  8  times  9  shillings  =  72  shilling-s,  and  4  shillings 
from  the  former  product  added,  are  76  shillings,  which,  since  20  s.  = 
£1,  must  equal  as  many  pounds  as  there  are  times  20  in  76,  which  are 
3  times,  with  a  remainder  of  16.     Hence,  76  s.  =  £3  16  s. 

Writing  the  16  s.,  and  reserving  theX3  to  add  with  the  pounds  of  the 
next  product,  we  have  8  times  £3  =  £24,  and  £3  added,  =  £27^  which, 
being  the  last  product,  we  write. 


78.    Methods  of  Proof. 

First  Method.  —  Go  over  the  work  a  second  time  in  the 
same  manner  as  before. 

Second  Method.  —  Consider  the  multiplicand  as  the  multi- 
plier, and  see  if  this  gives  the  same  result  as  before. 

The  figures  being  in  this  way  presented  in  a  different  order,  we  shall 
not  be  liable  to  repeat  any  mistake  we  may  have  made  in  the  first  work. 

Third  Method.  —  Write  out  by  itself  the  product  of  the 
multiplication  of  each  denomination,  begiiming  either  at  the 
left  or  right,  and  afterwards  add  these  products  together. 
The  sum  should  equal  the  former  product. 

Below  is  the  written  work  of  the  examples  in  VB  and  "77,  as  proved 
by  beginning  at  the  left,  and  writing  each  denomination  of  the  product 
separately. 

Exainple  1.      75.69  X  7 

490.      =  Product  of  70  by  7. 
35.      =  Product  of  5  by  7. 
4.2    ==  Product  of  .6  by  7. 
.63  =  Product  of  .09  by  7. 

529.83  =  Product  of  75.69  by  7  =  former  Product 
Example  2. 
£  3        9  s.        7  d.  X  8 


£24  =  £24  =  Product  of  £3  by  8. 

72  s.  =£3128.         =  Product  of  9  s.  by  8 

56  d.         =  4  s.  8  d.  =  Product  of  7  d.  by  8. 

£24       72  s.       56  d-         =  £27  16  s.  8d.  =  Product  of  £3  9  s.  7  d 

by  8  =  former  Product. 

Fourth  Method.  —  Another  method  of  proof  is,  after  hav- 


82  MUI.TII'I.ICATION. 

ing  written  out  the  work  as  at  first  performed,  to  begin  at  the 
left  hand,  thus  :  — 

75.69 
7 


529.83 
7  times  7  tens  are  49  tens ;  but  as  there  are  52  tens  in  the  product,  3 
tens  must  have  come  from  the  product  of  the  lower  denominations.  3 
tens  =  30  units,  and  adding  to  this  the  9  units  written  in  the  unit* 
place,  we  find  there  ought  to  be  39  units  in  the  product.  7  times  5  units 
are  only  35  units ;  hence,  if  the  work  be  right,  4  units  must  have  come 
from  the  product  of  the  lower  denominations.  4  units  =  40  tenths,  and 
adding  to  this  the  8  tenths  written  in  the  tenths'  place  of  the  product, 
we  find  that  there  ought  to  be  48  tenths  in  the  product.  7  times  6  tenths 
are  only  42  tenths ;  hence,  if  the  product  be  right,  6  tenths  must  have 
come  from  the  product  of  the  hundredths.  €  tenths  =  60  hundredths, 
and  adding  to  this  the  3  hundredths  written  in  the  hundredths'  place,  we 
find  that  there  ought  to  be  63  hundredths  in  the  product ;  and  as  7  times 
9  hundredths  are  63  hundredths,  we  may  infer  that  the  work  is  correct. 

£.       s.     d. 

3        9      7  ■ 

8 


27     16    8 


8  times  £3  =  £24.  But  as  in  the  written  product  there  are  £27, 
£3  must  have  come  from  the  lower  denominations.  £3  =  60  s.,  and 
adding  to  this  the  16  8.  written  in  the  shillings  of  the  product,  we  find 
that  there  ought  to  be  76  shillings  in  the  product.  8  times  9  s.  =  72  s. ; 
therefore,  if  the  work  be  right,  4  shillings  must  have  come  from  the 
lower  denominations.  4  s.  =  48  d.,  and  adding  to  this  the  8  d.  already 
written,  we  find  there  ought  to  be  56  d.  in  the  product;  and  as  8  times 
7  d.  =  56  d.,  we  infer  that  the  work  is  correct. 

If  the  computer  should  find  by  multiplying  numbers  in  one  method  a 
result  diflferent  from  that  obtained  by  multiplying  the  same  numbers  in 
some  other  method,  he  may  be  sure  that  there  is  an  error  in  one  opera- 
tion or  the  other,  and  he  should  examine  his  work  patiently  till  he  finds 
it.  No  person  who  is  willing  to  allow  an  error  to  pass  undetected  can 
be  a  good  arithmetician.     (See  54.) 

70.    Problems.- — Multiplier  a  single  Figure. 

NoTB.  —  This  article  includes  reduction  descending.  (See  Note 
tftor  36th  example.) 


MULTIPLICATION.  83 

1.  What  is  the  product  of  84687  X  4  ? 

2.  What  is  the  product  of  .0078673  X  7  ? 

3.  What  is  the  product  of  237.904  X  8  ? 

4.  What  is  the  product  of  20078  X  9  ? 

5.  What  is  the  product  of  .00978  X  6  ? 

6.  What  is  the  product  of  796.783  X  7  ' 

7.  What  is  the  product  of  .00978  X  6  ? 

8.  What  is  the  product  of  37842  X  8  ? 

9.  What  is  the  product  of  .7948  X  8  ? 

10.  1  pound  Avoirdupois  of  distilled  water  contains 
27.7015  cubic  inches.  How  many  cubic  inches  will  8  pounds 
contain  ? 

Reasoning  Process.  —  If  1  potrad  contains  27.7015  cubic  inches,  8 
pounds  will  contain  8  times  27.7015  cubic  inches,  wliich  can  be  found  by 
multiplying  it  by  8. 

11.  What  will  7  acres  of  land  cost  at  $94,839  per  acre  ? 

12.  1  pound  Troy  of  distilled  water  contains  22.794377 
cubic  inches.     How  many  cubic  inches  will  8  pounds  contain? 

13.  How  many  cubic  inches  are  there  in  7  cubic  feet? 

14.  How  much  will  it  cost  to  build  9  miles  of  railroad  at 
$19783.27  per  mile? 

15.  How  much  will  3  farms  cost  at  $3879.86  each  ? 

1 6.  How  many  feet  would  a  man  walk  in  6  days  at  the 
rate  of  56487  feet  per  day  ? 

17.  How  many  miles  would  a  ship  sail  in  9  weeks,  if  she 
sails  at  the  rate  of  1198.47  miles  per  week  ? 

18.  How  many  inches  are  there  in  4  miles,  there  being 
63360  inches  in  1  mile  ? 

19.  How  many  pounds  are  there  in  5  loads  of  hay,  each 
weighing  2794  pounds  ? 

20.  How  many  acres  in  7  lots,  each  containing  24.74386 
acres  ? 

21.  How  many  square  rods  in  a  road  754  rods  long  and  4 
rods  wide  ? 

Reasoning  Process.  —  Since  a  space  1  rod  long  and  1  rod  wide  con- 
tains 1  square  rod,  a  space  754  rods  long  and  1  rod  wide  must  contain 


84  MULTIPLICATION. 

754  square  rods,  and  a  space  754  rods  long  and  4  rods  wide  must  con 
tain  4  times  754  square  rods,  which  may  be  found  by  multiplying  754 
by  4.    (See  40,  Note.) 

22.  How  many  square  feet  in  a  walk  7&6  feet  long  and  9 
feet  wide  ? 

23.  How  many  square  feet  in  a  wall  437  feet  long  and  6 
feet  high  ? 

24.  Mr.  Haven's  garden  is  124  feet  long  and  97  feet  wide, 
and  is  enclosed  by  a  tight  board  fence  5  feet  high.  How 
many  square  feet  of  boards  are  there  in  the  fence  ? 

Suggestion  to  the  Student.  —  Draw  a  plan  of  the  garden. 

25.  Mr.  Haven  laid  out  a  gravel  walk,  4  feet  wide,  within 
the  fence,  and  extending  all  around  the  garden.  How  many 
square  feet  did  it  contain  ? 

Suggestion  to  the  Student.  —  Draw  a  plan  of  the  garden  and  walk. 

26.  What  will  49.67  barrels  of  apples  cost  at  $3  per 
barrel  ? 

Reasoning  Process.  —  If  1   barrel  costs  3  dollars,  49.67  barrels  will 
cost  49.67  times  3  dollars,  which  is  equivalent  to  3  times  49.67  dollars. 
The  work  would  be  written  thus  :  — 

$49.67  =  Multiplicand. 
3  =  Multiplier. 


$149.01  =  Product. 

Another  Reasoning  Process.  —  It  is  evident  that  if  each  barrel  should 
eost  a  dollar,  all  would  cost  as  many  dollars  as  there  are  barrels  bought, 
which  in  this  instance  would  be  $49.67.  But  if,  at  $1  per  bbl.,  they  cost 
$49.67,  at  $3  per  bbl.,  they  would  cost  3  times  $49.67. 

The  work  may  be  written  thus  :  — 

49.67  bbl.  at  $3  per  bbl. 

$  49.67  =  cost  at  $1  per  bbl. 
3  times  $49.67  =  $149.01  =  cost  at  $3  per  bbl.  =  Ans. 

27.  What  will  3749  lb.  of  saleratus  cost  at  .08  per  lb.  ? 

28.  What  will  178.69  bbl.  of  flour  cost  at  $7  per  bbl.  ? 

29.  What  will  27.96  firkins  of  butter  cost  at  $9  per  firkin? 

30.  What  will  be  the  weight  of  3794  cannon  balls,  each 
ball  weighing  8  lb  ? 


MULTIPLICATION.  85 

81.  If  a  soldier  eats  4  lb.  of  meat  in  a  week,  how  many 
pounds  will  2896  soldiers  eat  in  the  same  time  ? 

32.  What  will  4736  casks  of  raisins  cost  at  $7  per  cask  ? 

33.  If  a  vessel  sails  uniformly  at  the  rate  of  9  miles  per 
hour,  how  far  wiU  she  sail  in  476  hours  ? 

34.  How  many  bushels  in  1487  barrels,  if  each  barrel 
holds  3  bushels  ? 

35.  How  much  will  3479  window  weights  weigh,  if  each 
weighs  6  pounds  ? 

36.  82  bu.  3  pk.  5  qt.  1  pt.  =  how  many  pints  ? 

Reasoning  Process.  —  Since  there  are  4  pecks  for  every  bushel,  there 
must  be  4  times  as  many  pecks  as  bushels,  or,  in  this  case,  4  times  82 
pecks,  which  are  328  pecks,  and  adding  3  pecks  to  this  gives  331  pecks 
as  the  value  of  82  bu.  3  pk. 

Since  there  are  8  quarts  for  every  peck,  there  must  be  8  times  as 
many  quarts  as  there  are  pecks,  or,  in  this  case,  8  times  331  quarts, 
which  are,  &c. 

Another  Form.  —  Since  1  bushel  =  4  pecks,  82  bushels  must  equal 
82  times  4  pecks,  equivalent  to  4  times  82  pecks,  which  are  328  pecks, 
and  adding  3  pecks  to  this  gives  331  pecks,  as  the  value  of  82  bu.  3  pk. 

Since  1  peck  =  8  quarts,  331  pecks  must  equal  331  times  8  quarts, 
equivalent  to  8  times  331  quarts,  which  are,  &c. 

Note.  —  Questions  like  the  above,  requiring  that  the  value  of  num- 
bers of  one  denomination  shall  be  expressed  in  terms  of  some  lower 
denomination,  are  called  questions  in  Reduction  Descending;  but  as 
they  do  not  differ  at  all  from  other  questions  in  multiplication,  they 
require  no  separate  treatment.  In  performing  them,  there  is  no  need  of 
writing  the  multipliers,  as  they  may  be  known  from  the  table  of  the 
weight  or  measure  used.  In  reducing  to  any  denomination  of  which 
there  are  units  already  expressed,  it  will  usually  be  more  convenient  to 
add  those  units  at  the  time  we  make  the  multiplication. 

METHOD    OF    WRITING   THE    WORK. 

82  bu.  3  pk.  5  qt.  1  pt. 
331  pk.  =  82  bu.  3  pk. 


2653  qt.  =  82  bu.  3  pk.  5  qt. 
5307  pt.  =  82  bu.  3  pk.  5  qt.  1  pt. 
8 


86  MULTIPLICATION. 

The  following  abbreviated  form  may  be  adopted  after  tb« 
above  is  perfectly  familiar :  — 

82  bu.  3  pk.  5  qt.  1  pt. 


331  pk. 
2653  qt. 
5307  pt.  =  Ans, 

37.  97  bu.  2  pk.  7  qt.  1  pt.  =  how  many  pints  ? 

38.  187  bu.  1  pk.  6  qt.  0  pt.  3  gi.  =  how  many  gills  ? 

39.  238  gal.  3  qt.  1  pt.  2  gi.  =:  how  many  gills  ? 

40.  596  gal.  2  qt.  1  pt.  3  gi.  =:  how  many  gUls  ? 

41.  91b  lis  63  23  =  how  many  scruples  ? 

42.  8  lb  3§  7 3  1  9  =  how  many  scruples  ? 

43.  937  le.  1  m.  5  fur.  =  how  many  furlongs  ? 
*»    44.  286  le.  2  m.  7  fur.  =:  how  many  furlongs  ? 

45.  What  will  37  bu.  2  pk.  3  qt.  1  pt.  of  cherries  cost  at  4 
cents  per  pint  ? 

46.  What  will  114  bu.  3  pk.  2  qt.  of  wheat  cost  at  1  cent 
per  gill  ? 

47.  What  will  17  gal.  3  qt.  1  pt.  3  gi.  of  oil  cost  at  5  cents 
per  gill  ? 

48.  What  will  93  gal.  2  qt.  1  pt.  2  gi.  of  brandy  cost  at  8 
cents  per  gill  ? 

49.  What  is  the  product  of  £22  18  s.  8  d.  2  qr.  by  7  ? 

£.       8.         d.    qr. 

22     18       8     2  =  Multiplicand. 
7  =  Multiplier. 


160     10     11     2  =  Product. 

Explanation.  —  7  times  2  qr.  =  14  qr.,  which  (since  4  qr.  =«  1  d.,) 
equal  as  many  pence  as  there  are  times  4  in  14,  which  are  3  times,  with 
a  remainder  of  2.    Hence,  14  qr.  =  3  d.  2  qr. 

7  times  8  d.  =  56  d.,  and  3  d.  added  =  59  d.,  which  (since  12  d.  =s 
1  8.)  equal  as  many  shillings  as  there  are  times  12  d.  in  59  d.,  which  are 
4  times,  with  a  remairider  of  11.    Hence,  59  d.  =  4  s.  11  d. 

7  times  ISs.  s^  126  s.,  and  4  s.  added  =  130  •.,  which  (since  20  s 


MULTIPLICATION.  87 

■=  £1)  equal  as  many  pounds  as  there  are  times  20  in  130,  which  are  6 
times,  with  a  remainder  of  10.    Hence,  130  s.  =  £6  10  s. 

7  times  £22  =  £154,  and  £6  added  =  £160. 

Hence,  the  product  is  £160  10  s.  lid.  2  qr. 

Note.  —  When  any  denomination  to  be  multiplied  is  very  near  a 
unit  of  the  next  higher,  the  work  may  frequently  be  much  shortened  by 
considering  it  a  unit  of  that  higher  denomination,  and  subtracting  for 
its  deficiency  in  value. 

For  instance,  in  the  example  above:  since  18s,  =  £1  —  2s.,  7 
times  18  s.  must  equal  £7  —  14  s.,  or  £6  6  s.,  to  which  adding  4  s.,  (from 
the  previous  product,)  we  have  £6  10  s.,  as  before. 

What  is  the  product  — 

50.  Of  £29  8  s.  11  d.  1  qr.  multiplied  by  9  ? 

51.  Of  37T.  19cwt.  2qr.  241b.  11  oz.  7  dr.  multipUed 
bj3? 

52.  Of  273  bu.  1  pk.  5  qt.  1  pt.  multiplied  by  2  ? 

53.  Of  9  lb,  8  oz,  13  dwt.  22  gr.  multiplied  by  7  ? 

54.  Of  28  da.  17  h.  27  m.  58  sec.  multiplied  by  6  ? 

55.  Of  47tb  8§  73  29  18  gr.  multiplied  by  4  ? 

56.  Of  238  gal,  1  qt.  1  pt,  3  gi,  multiplied  by  9  ? 

57.  Of  674  lb.  4  oz.  19  dwt.  20  gr.  multiplied  by  8  ? 

58.  Of  23  lb.  4  oz.  16  dwt.  22  gr.  multiplied  by  8  ? 

59.  Of  13  cwt.  2  qr.  17  lb.  13  oz.  9  dr.  multiplied  by  6  ? 

60.  Of  6  T.  18  cwt  1  qr.  24  lb.  2  oz.  1  dr.  multipUed  by  7  ? 

61.  Of  9  bu.  3  pk.  7  qt,  multiplied  by  9  ? 

62.  Of  9  gal.  2  qt.  1  pt.  multiplied  by  5  ? 

63.  Of  8  w.  1  da,  23  h.  59  m.  56  sec.  multiplied  by  7  ?  * 

64.  Of  £8  19  s.  11  d.  3  qr.  multiplied  by  8  ? 

65.  Of  9  T.  19  cwt.  3  qr.  24  lb,  14  oz.  multiplied  by  7  ? 

66.  Of  7  lb.  11  oz.  19  dwt.  21  gr.  multiplied  by  4  ? 

67.  Of  483  yd.  3  qr.  1  na.  multiplied  by  9  ? 

68.  Of  4978  bu.  3  pk,  5  qt.  multiplied  by  5  ? 

69.  Of  37  lb.  11  oz.  19  dwt,  23  gr.  multiplied  by  6  ? 
70     Of  £5871   18  s.  4  d.  1  qr.  multiplied  by  2  ? 


*  In  performing  this  example,  much  labor  may  be  saved  by  observing 
that  the  multiplicand  is  only  4  seconds  less  than  8  w.  2  da.  Similar  thinga 
can  frequf  ntly  be  applied,  as  in  several  of  the  subsequent  examples. 


88 


MULTIPLICATION. 


80.    Multiplication  by  Factors. 

(a.)  It  often  happens  that  a  number  used  as  a  multiplier  is 
the  product  of  two  or  more  factors.  In  such  cases  it  is 
sometimes  convenient  to  resort  to  processes  similar  to  those 
explained  below. 

Note.  —  In  writing  the  work  here,  as  in  several  other  places  through- 
out the  book,  we  have  used  letters  for  convenience  of  indicating  to  the 
eye  the  operations  which  have  been  performed,  and  the  relations  which 
the  numbers  bear  to  each  other.  For  instance,  in  the  first  four  given 
below^  "  a  =  743,"  means  that  the  letter  "  a"  stands  for  743.  "  2  X  a  = 
b  =  1486,"  means  that  two  times  the  number  "  a,''  i.  c.,  2  times  743,  =« 
the  number  represented  by  *'b,"  which  is  1486.  "  6  X  b  =  12  X  a,' 
means  that  6  times  the  number  "  b  "  (i.  e.,  6  times  1486)  equals  12  times 
the  number  "a,"  (i.  e.,  12  times  743.) 

The  student  will  observe  that  the  letter  which  in  one  form  stands 
for  one  number,  may  in  another  form  stand  for  a  different  number. 
Thus,  in  the  first  form  "b"  is  used  to  represent  1486,  while  in  the 
Becond  it  is  used  to  represent  2972. 

How  many  are  12  times  743  ? 


FIRST   METHOD. 

a  == 

2  X  a  =  b  = 

6Xb  =  12Xa  = 


743 
2 

1486 
6 

8916 


THIRD    METHOD. 


a=    743 
3 


3  X  a  =  b  =  2229 
2 


SECOND 

METHOD 

a  = 

743 
4 

4  X  a 

X  ar= 

2972 
3 

b=  12 

8916 

FOURTH 

METHOD. 

a  = 

743 
2 

2  X  a 

1486 
2 

2Xb=:6Xa  =  c~  4458       2Xb=4Xa  =  c=:  2972 
2  3 


2Xc=12Xa  =  8916  3Xc=12Xa  =  8916 


MULTIPLICATION.  89 

Explanations. 

First  Method.  —  Since  12  =  6  times  2,  12  times  a  number  must  be 
equal  to  6  times  2  times  tbe  number. 

Second  Method. —  Since  12  =  3  times  4,  12  times  a  number  must  be 
equal  to  3  times  4  times  the  number. 

Hiird  Method. —  Since  12  =  2  times  2  times  3,  12  times  a  number 
must  be  equal  to  2  times  2  times  3  times  the  number. 

Fourth  Method. —  Since  12  =  3  times  2  times  2,  12  times  a  number 
must  equal  3  times  2  times  2  times  the  number. 

(6.)    Solve  the  following  examples  in  a  similar  manner. 
What  is  the  val^e  — 


1.  Of  879  X  18? 

2.  Of  9874  X  27  ? 
3.,  Of  8764  X  36  ? 


4.  Of  6427  X  42  ? 

5.  Of  4.379  X  64? 

6.  Of  2976.4  X  28  ? 

7.  Of  2377  T.  17  cwt.  2  qr.  19  lb.  6  oz.  11  dr.  X  63  ? 

8.  Of  271b  8i  65  29  17 gr.  X  45? 

9.  Of  19  w.  5  d.  17  h.  38  m.  29  sec.  X  36  ? 

10.  Of  £28  13  s.  10  d.  2  qr.  X  35  ? 

11.  Of  48  lb.  10  oz.  16  dwt.  19  gr.  X  24  ? 

12.  Of  837  bu.  3  pk.  6  qt.  1  pt.  X  18  ? 

(c.)  The  most  useful  application  of  the  foregoing  principle 
is  made  when  the  multiplier  is  some  multiple  of  10. 

13.  What  is  the  product  of  8746  X  400  ? 

Solution.  —  Since  400  =  4  times  100,  400  times  a  number  must  equal 
4  times  100  times  the  number;  to  obtain  which  we  have  only  to  remove 
the  point  two  places  towards  the  right,  (24)  and  multiply  by  4.   Hence, 

8746 
400 


3498400 

14.   What  is  the  product  of  9.7487  X  7000  ? 

Solution.— ^mc^  7000  =  7  times  1000,  7000  times  a  number  must 
equal  7  times  1000  times  the  number;  to  obtain  which  we  have  only  to 
multiply  by  7,  and  remove  the  point  three  places  towards  the  right. 
Hence. 

9.7487 

7000 


Y 


68240.9 
8   ♦ 


90  MULTIPLICATION. 

(d.)  In  like  manner,  to  multiply  by  60,  we  may  multiply 
by  6,  and  remove  the  point  one  place  towards  the  right ;  to 
multiply  by  9000000,  we  may  multiply  by  9,  and  remove  the 
point  six  places  towards  the  right.  In  any  case,  all  places 
left  vacant  between  the  number  and  the  point  must  be  filled 
with  zeros.     (See  15.) 


What  is  the  product  — 

15.  Of  874379  X  20  ? 

16.  Of  27.9863  X  5000? 

17.  Of  714.26  X  90000? 

18.  Of  62.794  X  40  ? 


19.  Of  627.34  X  80? 

20.  Of  9137.6  X  30000? 

21.  Of  84273  X  60? 

22.  Of  7*643  X  7000000? 


81.     When  both  Factors  are  large  Numbers,  , 

(a.)  We  can  find  the  product  of  two  numbers  by  multiply- 
ing one  of  them  by  the  parts  into  which  we  choose  to  separate 
the  other,  and  then  adding  the  products  thus  obtained  together. 

Illustration.  —  8  times  7=6  times  7  +  2  times  7  =  3  times  7  -f"  5 
times  7  =  7  times  7  +  once  7=4  times  7  -f-  4  times  7  =  56. 

{b.)  This  principle,  and  the  one  illustrated  in  article  80, 
are  ordinarily  employed  when  the  multiplier  contains  several 
denominations. 

Illustrations.  —  We  usually  get  83  times  a  number,  by  adding  together 
80  times  the  number  and  3  times  the  number. 

We  get  647  times  a  number  by  adding  together  600  times  the  num- 
ber, 40  times  the  number,  and  7  times  the  number. 

Wc  get  8009  times  a  number  by  adding  together  8000  times  the  num- 
ber and  9  times  the  number. 

(c.)  It  can  make  no  difference  which  part  of  a  number  we 
muhiply  by  first,  provided  we  multiply  by  all  its  parts ;  yet 
for  the  sake  of  uniformity  it  may  be  well  generally  to  begin 
with  the  lowest  denomination. 

1.  Suppt)se  that  we  are  required  to  find  the  product  of 
5794  X  78? 

Explanation.  —  We  may  find  the  product  by  8  in  the  usual  manner. 
To  find  the  product  by  70,  we  have  only  to  multiply  by  7,  and  remove 
the  point  one  place  towards  the  right,  or,  which  is  the  same  thing,  the 


MULTIPLICATION.  91 

figures  one  place  towards  the  left.    Adding  these  results  together  will 
give  78  times  the  number. 

WRITTEN    WORK. 

a  =      5794  =  Multiplicand. 
78  =  Multiplier. 

8  X  a  =  b  =    46352  =  Product  by  8. 
70  X  a  =:  c  =  405580  =  Product  by  70. 

b  +  c==78Xai=  451932  =  Product  by  78. 

(d.)  Since  the  product  of  the  multiplication  by  the  units  is 
sufficient  to  fix  the  place  of  the  figures  in  the  subsequent 
products,  the  zero  at  the  right  of  the  second  product  need  not 
have  been  written.     The  product  would  then  stand,  — 


4345  =  Product  by    5. 
6083    =1        "         "    70. 

65175  =       ^'        "    75. 

(e.)    Examples  in  which  the  multiplier  consists  of  more 
than  two  figures  are  performed  in  a  similar  way. 
2.   What  is  the  product  of  780.69  X  20850  ? 

WRITTEN    WORK. 

a  =z  780.69  =  Multiplicand. 

20850  =  Multiplier. 

a  X  50  =  b  ==    39034.50  =  Product  by  50. 

a  X  800  =  c  =   624552.   =  Product  by  800. 

a  X  20000  =  d  =  156138     =  Product  by  20000. 

b  +  c  +  d  =  16277386.50  =  Product  by  20850. 

Note.  —  In  practice,  only  the  necessary  figures  should  be  written 
Thus :  — 

780.69 
20850. 


39034.50 
624552. 
1.56138 

16-277386..50 


What,  is  the  product  of — 

3. 

Of  80276  X  39  ? 

13. 

4. 

Of  298794  X  148  ? 

14. 

5. 

Of  273986  X  27  ? 

15. 

6. 

Of  4943  X  78  ? 

16. 

7. 

Of  23879  X  2741  ? 

17. 

8. 

Of  84976  X  203  ? 

18. 

9. 

Of  25873  X  506  ? 

19. 

10. 

Of  47.296  X  37  ? 

20. 

11. 

Of  423758  X  6200  ? 

21. 

12. 

Of  594.27  X  30200  ? 

22. 

92  MULTIPLICATION. 


Of  30678  X  427  ? 

Of  5.796  X  238? 

Of  45.718  X  432? 

Of  .00876  X  74  ? 

Of  67.968  X  327  ? 

Of  970062  X  37  ? 

Of  29743  X  806  ? 

Of  8427  X  3076  ? 

Of  279437  X  27623  ? 

Of  8.64298  X  43000  ? 

(/.)    It  is  obvious  that  the  product  obtained  by  muhiplying 

one  number  by  the  difference  of  two  numbers,  is  equivalent 

to  the  difference  of  the  products  obtained  by  multiplying  the 

numbers  separately  by  the  two  numbers. 

Illustration.  —  5  times  3  =  7  times  3  —  2  times  3  =  8  times  3  —  3 
times  3  =  15  times  3  —  10  times  3  =  29  times  3  —  24  times  3,  &c 

(g.)  This  principle  is  the  reverse  of  that  stated  in  (a,)  and 
can  often  be  advantageously  applied,  as  illustrated  below. 

{h.)  To  multiply  by  99,  observe  that  since  99  =  100  —  1,  99  times  a 
number  must  equal  100  times  the  number  minus  once  the  number.  For 
example,  — 

99  times  837  =  100  times  837  —  837  =  83700  —  837  =  82863. 

(i.)  To  multiply  by  999,  observe  that  since  999  =  1000  —  1,  999 
times  a  number  must  equal  1000  times  the  number  minus  once  the  num- 
ber.   For  example,  — 

999  times  14.67  =  1000  times  14.67  —  14.67  =  14670  —  14.67  == 
14655.33. 

{j.)  To  multiply  by  699,  observe  that  since  699  =700  —  1,  699 
^iipes  a  number  must  equal  700  times  the  number  —  once  the  number 
Ft.r  example, 

699  X  5784  =  700  X  5784  —  5784  =  4048800  —  5784  =  4013016 

{h)    In  like  manner  we  should  have  — 

49  X  785  =  50  X  785  —  785. 

98  X  4697  =  100  X  4697  —  2  X  4697. 

79996  X  394845  =  80000  X  394845  —  4  X  394845. 

"What  is  the  product  — 

23.   Of  48673  X  29?  |     24.   Of  37848  X  99? 


MULTIPLICATION.  93 


25.  Of  69435  X  69  ? 

26.  Of  29485  X  999  ? 

27.  Of  7486  X  998  ? 

28.  Of  4278  X  3999  ? 


29.  Of  6786  X  49  ? 

30.  Of  4296  X  79  ? 

31.  Of  28643  X  999  ? 


82.    Abbreviated  Method, 

(a.)  When  the  multiplier  consists  of  more  than  one  de- 
nomination, much  labor  in  writing  figures  may  be  saved  by 
applying  the  principles  illustrated  in  the  following  exam- 
ples :  — 

1.    What  is  the  product  of  8356  multiplied  by  79  ? 

Preliminary  Explanation.  —  It  is  evident  that,  in  performing  the 
required  multiplication,  we  shall  obtain  units  by  multiplying  6  units  by 
9  units.  We  shall  obtain  tens  by  multiplying  5  tens  by  9,  and  6  units 
by  7  tens,  and  we  may  have  some  from  the  product  of  the  units.  We 
shall  obtain  hundreds  by  multiplying  3  hundreds  by  9,  and  5  tens  by  7 
tens,  and  we  may  have  some  from  the  former  products.  We  shall  obtain 
the  other  denominations  in  a  similar  manner.  We  may  then  proceed 
thus,  writing  the  figures  of  each  denomination  as  usual :  — 

WRITTEN    WORK. 

8356     Multiplicand. 
79     Multiplier. 

660124     Product 

Explanation.    9  times  6  units  =  54  units.    We  write  4  units. 

9  times  5  tens  =  45  tens,  +  5  tens  (from  the  product  of  the  units)  =a 
50  tens,  +  7  tens  times  6,  or  42  tens,  =  92  tens  =  9  hundreds  and  2 
tens.    We  write  2  tens. 

9  times  3  hundreds  =  27  hundreds,  -f-  9  hundreds  (from  the  previous 
product)  =  36  hundreds,  +  7  tens  times  5  tens,  or  35  hundreds,  =  71 
hundreds  =  7  thousand  and  1  hundred.     We  write  1  hundred. 

9  times  8  thousands  =  72  thousands,  -f-  7  thousands  (from  the  pre- 
vious product)  =  79  thousands,  -(-  7  tens  times  3  hundreds,  or  21  thou- 
sands, =  100  thousands  =  10  ten-thousands.  We  write  0  in  the  thou- 
sands' place  of  the  product. 

7  tens  times  8  thousands  =  56  ten-thousands,  +  10  ten-thousands 
(from  the  previous  product)  =  66  ten-thousands,  which  we  write. 

The  multiplication  being  now  completed  shows  the  answer  to  be 
660124. 


94  MULTrPLICATION. 

(h.)  The  following  exhibits  the  necessary  operations  on 
the  numbers,  and  is  practically  a  much  more  convenient  solu- 
tion than  the  full  form  above  given. 

9  X  6  =  54  units. 

5  (from  last  product)  -f9X  5, +  7X6  =  54- 45 -f  42  =  92  tens. 

9  (from  last  product)  4-9  X  3,-1-7  X  5  =  9-}- 27 -f- 35  =  71  hun- 
dreds. 

7  (from  last  product)  -f-9  X  8,  +  7  X  3  =  74- 72  4-21  =  100 
thousands. 

10  (from  last  product)  4-  7  X  8  =  10  4"  56  =  66  ten-thousands. 
This  gives  for  an  answer  660124,  as  did  the  first  method. 

(c.)  The  last  process  being  understood,  the  work  may  be 
BtiU  further  abbreviated  by  omitting  to  name  the  factors  used. 

Thus,  54  units  =  5  tens  and  4  units. 

45  4-  5  =  50,  4-  42  =  92.     92  tens  =  9  hundreds  and  2  tens. 

27  4-  9  =  36,  4-  35  =  71.  71  hundreds  =  7  hundreds  and  1  thou- 
sand. 

72  4-  7  =  79,  4"  21  =  100.  100  thousands  =  10  ten-thousands  and 
0  thousands. 

56  4-  10  =  66.    66  ten-thousands. 

Answer,  as  before,  660124. 

(d.)  Finally,  the  work  may  be  abbreviated  so  as  to  name 
only  results :  — 

54  units  =  5  tens  and  4  units. 

45,  50,  92  tens  =  9  hundreds  and  2  tens. 

27,  36,  71  hundreds  =  7  thousands  and  1  hundred. 

72,  79,  100  thousands  =  10  ten-thousands  and  0  thousands. 

56,  66  ten-thousands. 

Note.  —  The  above  methods  are  much  more  expeditious  than  is  the 
method  of  writing  the  product  by  each  figure  of  the  multiplier  separate- 
ly, and  are  no  more  liable  to  inaccuracy. 

2.   How  much  will  97  acres  of  land  cost  at  $347  per  acre? 
3    If  a  cubic  yard  of  sand  weighs  2537  lb.,  how  much  will 
88  cubic  yards  weigh  ? 

4.  How  many  pounds  are  there  in  18  T.  17  cwt.  1  qr.  ? 

5.  If  a  ship  sails  96  miles  in  one  day,  how  far  will  she 
iail  in  247  days  ? 


MULTIPLICATION.  9ft 

6.  Bought  24  bundles  of  hay,  each  bundle  containing  497  lb. 
How  many  pounds  were  there  in  all  ? 

7.  Bought  2947  gallons  of  oil  at  $.84  per  gallon.  How 
much  did  it  cost  ?  Sold  it  for  $.97  per  gallon.  How  much 
was  received  for  it  ?     What  was  the  gain  on  it  ? 

8.  Mr.  Russell  bought  86  balls  of  twine,  each  ball  contain- 
ing 8794  ft.,  and  Mr.  Greene  bought  57  times  as  much.  How 
many  feet  of  twine  did  Mr.  Russell  buy  ?  How  many  did 
Mr.  Greene  buy  ? 

9.  How  much  will  83  casks  of  old  wine  cost  at  $138.47 
per  cask  ? 

10.  How  much  will  67  tons  of  lead  cost  at  $139.48  per 
ton? 

11.  Mr.  Hovey  bought  6247  feet  of  land,  and  Mr.  Ewell 
bought  94  times  as  much.  How  many  feet  did  Mr.  Ewell 
buy? 

12.  How  many  pounds  are  there  in  958  boxes  of  sugar, 
each  box  containing  743.67  lb.  ? 

Note.  —  The  products  and  sums  employed  in  solving  the  above 
example  are  given  below,  but  the  pupil  should  be  prepared  to  give  a 
more  full  explanation. 

56  hundredths  =  5  tenths  and  6  hundredths. 

5  -f-  48  +  35  =  88.     88  tenths  =  8  units  and  8  tenths. 

8  +  24  4-  30  -f-  63  =  125.     125  units  =  12  tens  and  5  units. 

12  +  32  -}-  15  +  54  =  113.     113  tens  =  11  hundreds  and  3  tens. 

11  4-  56  4-  20  -I-  27  =  114.  114  hundreds  =  11  thousands  and  4 
hundreds. 

11  +  35  4-  36  =  82.  82  thousands  =  8  ten-thousands  and  2  thou- 
sands. 

8  4-  63  =  71.     71  ten-thousands. 

The  answer,  therefore,  is  712435.86  lb. 

13.  How  many  are  8795  times  96543  ? 

Note.  —  By  extending  the  principles  before  explained,  we  can  writ* 
the  final  product  at  once,  as  below. 

96543 
8795 


849095685 


n 


►                                              MULTIPLICATION. 

In  the  following  forms,  two  products  are  written :  — 
96543 
8795 

9171585  units         = 
8399241       hundreds  = 

95  X  96543. 
8700  X  96543. 

849095685                     = 

96543 
8795 

8795  X  96543. 

76751685  units          = 
772344        thousands  = 

:  795  X  96543. 
=  8000  X  96543. 

849095685                     = 

:  8795  X  96543. 

14.  If  a  cubic  foot  of  iron  weighs  486.25  lb.,  how  much 
will  347  cubic  ft.  weigh  ? 

15.  How  many  square  feet  are  there  in  a  rectangular  lot, 
4327  feet  long  and  249  feet  wide  ? 

16.  How  much  will  48  acres  of  land  cost  at  $23,968  per 
acre  ? 

17.  What  will  798  tons  of  hay  cost  at  $14,278  per  ton  ? 

18.  I  bought  287  bales  of  cloth,  each  bale  containing 
247.986  yards.     How  many  yards  did  they  all  contain  ? 

19.  How  many  square  inches  are  there  in  a  lot  247  ft 
long  and  187  ft.  wide? 

20.  What  will  47983  yards  of  cloth  cost  at  $2.83  per  yd.  ? 

21.  What  wiU  7894  bbl.  of  flour  cost  at  $6.37  per  bbl.  ? 

22.  How  many  solid  inches  in  5  C.  6  Cd.  ft.  12  cu.  fl.  1437 
cu.  in.  ? 

23.  How  many  dr.  in  18T.  16cwt.  1  qr.  141b.  6oz.  11  dr.? 

24.  A  grain  dealer  sold  287  bushels  of  wheat  at  $1,294 
per  bushel,  and  1479  bushels  at  $1,267  per  bushel.  What 
did  he  receive  for  it  ? 

25.  I  bought  48  yards  of  broadcloth  at  $3,875  per  yd., 
153  yards  of  doeskin  at  $1,166  per  yd.,  and  379  yards  of 
cassimere  at  $1,125  per  yd.   What  was  the  cost  of  the  whole? 

26.  Mr.  Aldrich  owns  4  house  lots,  the  first  328  ft.  long 
and  189  ft.  wide;  the  second  437  ft  long  and  249  it,  wide; 


DITISICW.  97 

the  third  129  ft.  long  and  88  ft.  wide;  and  the  fourth  97  ft. 
long  and  86  ft.  wide.  How  many  square  feet  are  there  in  all 
of  them  ? 

27.  Mr.  Whitney  sold  84  acres  of  land  at  $34.96  per  acre, 
138  acres  at  $27.58  per  acre,  and  427  acres  at  $49.64  per 
acre.     How  much  did  he  seU  the  whole  for  ? 

28.  A  city  merchant  went  into  the  country  to  purchase 
flour.  He  was  absent  from  the  city  27  days,  and  his  expenses 
while  absent  were  $7,386  per  day.  He  bought  175  bbl.  of 
flour  at  $5,875  per  bbl.,  516  bbl.  at  $5,948  per  bbl.,  1386  bbl. 
at  $6.11  per  bbl.,  and  827  bbl.  at  $6,087  per  bbl.  It  cost 
him  $.634  per  bbl.  to  get  the  flour  transported  to  the  city. 
He  sold  697  bbl.  of  it  at  $7,114  per  bbl.,  824  bbl.  at  $7,213 
per  bbl.,  and  the  rest  at  $6,978  per  bbl.  Did  he  gain  or  lose 
by  the  adventure,  and  how  much  ? 

29.  A  merchant  bought  49.5  cases  of  cassimere,  each  case 
containing  297  yd.,  at  $1.1875  per  yd.  It  cost  him  $.125  per 
case  to  have  the  cloth  removed  to  his  store,  and  $.045  per 
case  to  have  it  hoisted  into  his  loft.  One  case  of  the  cloth 
was  stolen  from  him  ;  he  sold  23  cases  at  $1,423  per  yd.,  and 
the  remainder  at  $1,357  per  yd.,  agreeing  to  deliver  it  at  a 
railroad  depot,  1  mile  from  his  store.  It  cost  him  $.158  per 
case  to  have  it  carried  to  the  depot.  Did  he  gain  or  lose  on 
the  cloth,  and  how  much  ? 


SECTION    VII. 
DIVISION. 

S3*    Definitions  and  Ulmtrations. 
(a.)   Division  is  a  process  by  which  we  ascertain 

THE   NUMBER    OF   PARTS    OP   A    GIVEN    SIZE    INTO    WHICH   A 
CIVEN    NUMBER    MAY    BE     SEPARATED,    OR    BY     WHICH    WE 
9 


98  DIVISION. 

ASCERTAIN  THE  NUMBER  OF  UNITS  THERE  WILL  BE  Df 
EACH  PART  OBTAINED  BY  DIVIDING  A  GIVEN  NUMBER 
rNTO    A    GIVEN    NUMBER    OF   EQUAL    PARTS. 

(b.)    The  following  are  questions  in  division  :  — 

1.  42  =  how  many  times  6  ? 

2.  What  is  the  value  of  54  divided  by  9  ? 
8.   What  is  the  value  of  35  -7-  5  ? 

4.   How  many  apples  at  3  cents  apiece  can  be  bought  for  24  cents  * 

5    What  is  ^  of  36  1 

6.   If  8  apples  cost  24  cents,  how  much  will  1  apple  cost  1 

(c.)  In  the  first  of  the  above  questions,  we  are  required  to  find  how 
many  6's,  or  parts  of  6  each,  there  are  in  42  ;  in  the  second,  how  many 
9's,  or  parts  of  9  each,  there  are  in  54 ;  in  the  third,  how  many  5's,  or 
parts  of  5  each,  there  are  in  35 ;  in  the  fourth,  how  many  3's,  or  parts 
of  3  each,  there  are  in  24  ;  while  in  the  fifth  we  are  required  to  find  how 
many  units  there  will  be  in  each  part  obtained  by  dividing  36  into  4 
equal  parts ;  and  in  the  sixth,  how  many  units  there  will  be  in  each  part 
obtained  by  dividing  24  into  8  equal  parts. 

(d.)  This  shows  that  there  are  two  classes  of  questions  in  division, 
viz. :  one  class  in  which,  knowing  the  number  to  be  divided,  and  the  num 
ber  of  units  which  each  part  is  to  contain,  we  are  required  to  find  the 
number  of  parts ;  and  one  in  which,  knowing  the  number  to  be  divided, 
and  the  number  of  parts  into  which  it  is  to  be  divided,  we  are  required 
to  find  how  many  units  there  will  be  in  each  part;  i.  e.,  we  are  required 
to  find  a  fractional  part  of  a  number.  Both  classes  of  questions  may 
be  solved  by  the  same  numerical  process,  though  in  their  solution  they 
require  different  reasoning  processes. 

(c.)  The  number  to  be  divided  is  called  the  dividend.  In 
the  first  class  of  questions  the  number  which  shows  the  size 
of  each  part,  and  in  the  second  class  that  which  shows  the 
number  of  parts,  is  called  the  divisor.  The  result  is  called 
the  quotient. 

Illustrations.  —  In  the  first  of  the  above  questions,  42  is  the  dividend, 
6  is  the  divisor,  and  the  answer,  7,  is  the  quotient. 

In  the  second,  54  is  the  dividend,  9  is  the  divisor,  and  the  answer,  6, 
is  the  quotient. 

In  the  fifth,  36  is  the  dividend,  4  is  the  divisor,  and  the  answer,  9,  is 
tiie  quotient. 

(/.)    The  fourth  example  would  be  solved  thus  :  — 

If  for  3  c^ts  one  a]>ple  can  be  bought,  for  24  cents  as  many  apples 


DIVISION.  99 

can  be  bought  as  there  are  times  3  cents  in  24  cents,  which  are  8  times. 
Therefore,  8  apples  can  be  bought  for  24  cents,  when  1  apple  can  be 
bought  for  3  cents.  Here,  24  is  the  dividend,  3  is  the  divisor,  and  the 
answer,  8,  is  the  quotient.  The  divisor  and  dividend  are  of  the  same 
denomination,  and  the  quotient  is  the  number  of  times  the  divisor  is 
contained  in  the  dividend,  or  the  number  of  parts  equal  to  the  divisor 
which  the  dividend  equals. 

(g.)    The  sixth  example  would  be  solved  thus  :  — 

If  8  apples  cost  24  cents,  1  apple  will  cost  i  of  24  cents,  which  is  3 
cents.    Hence,  1  apple  will  cost  3  cents,  if  8  cost  24  cents. 

Here,  24  is  the  dividend,  8  is  the  divisor,  and  the  answer,  3,  is  the 
quotient.  The  dividend  and  quotient  are  of  the  same  denomination, 
and  the  divisor  shows  the  number  of  times  the  quotient  is  taken  in  the 
dividend,  or  the  number  of  parts  equal  to  the  quotient  which  the  divi- 
dend equals. 

(A.)  To  examine  more  fully  the  nature  of  the  processes,  let  us  see  by 
what  methods  the  answers  to  the  fourth  and  sixth  questions  could  be 
obtained. 

[i.)  It  is  obvious  that,  to  determine  the  answer  to  the  fourth  ques- 
tion, we  must  find  how  many  parts  of  3  cents  each  there  are  in  24  cents  j 
for  each  such  part  is  the  price  of  1  apple. 

We  can  do  this  by  counting  24  cents  into  piles  of  3  cents  each,  and 
then  counting  the  number  of  piles  ;  —  by  finding  how  many  threes  must 
be  added  to  make  24  ;  —  or  by  finding  how  many  times  3  equal  24,  by 
our  knowledge  of  multiplication.     This  last  process  is  division. 

{/)  To  detennine  the  answer  to  the  sixth  question,  we  must  find 
how  many  cents  there  tvill  be  in  each  part  obtained  by  separating  24 
cents  into  8  equal  parts. 

We  can  do  this  by  laying  out  24  cents  into  8  equal  piles,  and  then 
counting  the  number  of  cents  in  each  pile:  —  or  by  finding,  by  our 
knowledge  of  multiplication,  what  number  must  be  taken  8  times  to 
equal  24 ;  or,  which  will  give  the  same  numerical  answer,  by  finding 
how  many  times  8  equal  24.     The  last  process  is  division.  "* 

Note.  —  It  will  be  seen,  that  in  the  first  class  of  questions,  the  divi- 
sor and  dividend  are  of  the  same  denomination,  and  that  the  quotient 
expresses  the  number  of  times  the  divisor  is  contained  in  the  dividend, 
or  the  number  of  parts  equal  to  the  divisor  which  must  be  taken  to  pro- 
duce the  dividend ;  while  in  the  second  class,  the  divisor  expresses  the 
number  of  parts  into  which  the  dividend  is  to  be  divided,  and  the 
quotient  expresses  the  number  of  units  in  each  part;  thus  making  the 
dividend  and  quotient  of  the  same  denomination. 

(jfc.)   Practically,  division  is  the  reverse  of  multiplication.    In  the 


I 


100  Divii5ioir. 

latter,  the  factors  are  given,  and  we  are  required  to  find  the  product 
while  in  the  former,  one  of  the  factors  and  their  product  are  given,  and 
we  are  required  to  find  the  other ;  or,  when  the  dividend  will  not  exactly 
contain  the  divisor,  one  factor  and  the  product  of  the  two  plus  the 
remainder  are  given,  and  we  are  required  to  find  the  other  factor  and 
the  remainder. 

{I.)  The  divisor  is  always  the  given  factor,  the  quotient  is  the  re- 
quired factor,  and  the  dividend  is  the  product,  or  the  product  plus  the 
remainder.     The  remainder  is  always  less  than  the  divisor. 

(m.)    The  following  examples  illustrate  this  :  — 

1.  "  3  times  8,"  is  a  question  in  multiplication.  The  factors  3  and 
8  are  given,  and  the  product,  24,  is  required. 

2.  "  How  many  times  8  =  24,"  or  "  24  -f-  8  ?  **  are  questions  in 
division,  in  which  the  factor  8,  and  the  product  24,  are  given,  and  the 
missing  factor,  3,  is  required. 

3.  "  X  of  24,  or  3  times  what  number  =  24  ?  "  are  questions  in 
division,  in  which  the  factor  3,  and  the  product  24,  are  given,  and  the 
missing  factor  is  reqiiired. 

4.  "  In  8  times  7,  plus  5,"  the  factors  8  and  7  are  given,  and  their 
product,  plus  5,  which  is  61,  is  required. 

5.  "  In  61  -7-  7,"  or  "  61  =  how  many  times  7,"  the  factor  7,  and 
the  sum  of  the  product,  and  remainder,  which  is  61,  are  given,  and  the 
other  factor  and  remainder  are  required. 

6.  "  In  J.  of  61,"  or  "  7  times  what  number  =  61,"  the  factor  7,  and 
the  sum  of  the  product,  plus  the  remainder,  are  given,  and  the  other  fac- 
tor and  remainder  are  required. 

84.    Methods  of  Proof. 

From  the  preceding  illustrations  it  is  evident,  — 

First.  That,  where  there  is  no  remainder,  the  divisor  mul- 
tiplied by  the  quotient  must  produce  the  dividend ;  and  that 
the  dividend  divided  by  the  quotient  must  produce  the  divisor. 

Second.  That  if,  when  there  is  a  remainder,  the  divisor  and 
quotient  be  multiplied  together,  and  the  remainder  be  added 
to  their  product,  the  result  will  equal  the  dividend. 

Third.  That  if  the  remainder  be  subtracted  from  the  divi- 
dend, and  the  remainder  thus  obtained  be  divided  by  the 
divisor,  the  result  will  equal  the  quotient ;  or,  if  it  be  divided 
by  the  quotient,  the  result  will  equal  the  divisor. 

Note.  —  Tho  remain  ler  should  always  be  less  than  the  divisor ;  foi» 


Divisio^r,  101 

if  it  is  not,  the  dividend  will  contain  the  divisor  more  times  than  is  indi- 
cated by  the  quotient  figure. 


85.    Examples.  —  Quotient  a  single  Figure. 

(a.)  How  many  oranges  at  7  cents  apiece  can  be  bought 
for  61  cents  ? 

Reasoning  Process.  —  If  for  7  cents  1  orange  can  be  bought,  as  many 
oranges  can  be  bought  for  61  cents  as  there  are  times  7  in  61. 

Explanation.  —  To  perform  the  necessary  division,  we  observe  that  as 
56  =  8  times  7,  61  must  equal  8  times  7,  with  a  remainder  of  5 ;  or, 
since  5  -^  7  =  f-,  61  =  8f  times  7. 

Hence,  the  quotient  is  8,  and  the  remainder  is  7,  or  the  complete 
quotient  is  8A,  which  shows  that  8  oranges  can  be  bought,  leaving  5 
cents  unused,  or  that  85.  oranges  can  be  bought  with  all  the  money. 

First  Proof.  —  See  if  the  price  of  8  oranges  at  7  cents  apiece,  added 
to  5  cents,  will  make  61  cents. 

Second  Proof.  —  Considering  that  85  oranges  are  bought,  see  if  they 
will  cost  61  cents  at  7  cents  apiece. 

(b.)  The  above  are  proofs  of  the  correctness  of  the  entire  work ;  the 
following  only  test  the  correctness  of  the  division. 

Tliird  Proof     8  times  7  =  56,  and  5  added  =  61  =  dividend. 

Fourth  Proof  8  times  7  =  56  ;  and  56  from  61  leaves  5  =  re- 
mainder. 

Fifth  Proof      5  from  61  =56,  and  56  -f-  7  =i  8  ==  (quotient. 

(c.)    The  work  may  be  written  thus  :  '^j  ; ''!  .'    '.'  . 

Dividend.    ''       .\   t     >'*';'•.  ^';  ■    * 
Divisor  7  )  61  —  5  =  Relnairid^i'. '     '^'  '''•  '  ' 

8  zzi  Quotient. 

(d.)  Or,  by  expressing  the  division  of  the  remainder,  by 
placing  it  in  the  form  of  a  fraction,  we  have,  — 

Divisor  7  )  61    z=  Dividend. 

8^  =  Quotient. 

In  the  first  form,  the  remainder  is  undivided,  and  is  of  the  same  de- 
nomination as  the  dividend.    It  is  placed  opposite  the  dividend,  with 
the  minus  sign  between,  to  indicate  that  all  the  dividend  except  that 
9* 


102 


BI VI  SIGN. 


has  been  divided.    Indeed,  it  might  have  been  subtracted  from  the  divi- 
dend without  affecting  the  quotient. 

In  second  form,  the  entire  dividend  is  divided.    Hence,  the  A  is  not  a 
remainder,  but  is  a  part  of  the  quotient. 

Note.  —  The  distinction  between  a  remainder  and  a  fractional  quo- 
tient should  be  carefully  observed. 

What  is  the  quotient  — 


1.   Of  37  - 

r4? 

7.  Of  43  -j-  7  ? 

2.   Of83- 

-10? 

8.   Of  26  -^  3? 

3.   Of  39  - 

-7? 

9.   Of  47  -r-  5  ? 

4.   Of  17  - 

r  2? 

10.   Of  48  — 9? 

5.   Of  86  - 

^9? 

11.   Of  75-^8? 

6.    Of  57  - 

r6? 

12.   Of  41  -^  7  ? 

13.  How  many  apples  at  3  cents  apiece  can  be  bought  for 
28  cents  ? 

14.  How  many  barrels  of  flour  at  $9  per  barrel  can  be 
bought  for  $62  ? 

15.  How  many  yards  of  broadcloth  at  $6  per  yard  can  be 
bought  for  $53  ? 

Note.  —  The  student  should  practise  on  examples  like  the  above, 
till  he  can  perform  them  with  ease  and  rapidity. 


:^i^>  j  I^ividend  a  large  Number. 

{cu)„  .Wfceii;  laFge  numbers  are  to  be  divided,  we  begin  with 
the  higiiest  d-ejicinihatNJns,  and  proceed  as  in  the  following 
example. 

1.  How  many  barrels  of  flour  at  $8  per  barrel  can  be 
bought  for  $58455  ? 

Reasoning  Process.  —  If  for  $8,  1  barrel  can  be  bought,  as  many  bar 
rels  can  be  bought  for  $58455  as  there  are  times  8  in  58455. 

WRITTEN    WORK. 

Dividend. 
Divisor  =  8  )  58455  —  7  =  Remainder, 


7306  =:  Quotient, 


DIVISION.  106 

Explanation  of  Process  of  Dividing.  8  is  contained  in  58  thousands 
7  thousand  times,  with  a  remainder  of  2  thousands,  (for  7000  times  S  =• 
56000,  and  2000  added  =  58000.)  We  therefore  write  7  as  the  thou- 
sands' figure  of  the  quotient. 

Reducing  the  2  thousands  remaining  to  hundreds,  and  adding  to 
them  the  4  hundreds,  gives  24  hundreds  to  be  next  divided.  8  is  con- 
tained in  24  hundreds  3  hundred  times,  (for  300  times  8  =  2400.)  We 
therefore  write  3  as  the  hundreds'  figure  of  the  quotient. 

As  8  is  not  contained  as  many  as  ten  times  in  5  tens,  we  write  0  as 
the  tens'  figure  of  the  quotient,  and  reduce  the  5  tens  to  units,  consider- 
ing them  with  the  5  units. 

8  is  contained  in  55  units  6  times,  with  a  remainder  of  7  units,  (for  6 
times  8  =  48,  and  7  added  =  55.)  We  therefore  write  6  as  the  units' 
figure  of  the  quotient. 

This  gives  7306  for  the  quotient,  and  7  for  the  remainder ;  or,  since 
7  -i-  8  =  7 .  it  gives  7306|.  for  a  quotient. 

Hence,  7306  barrels  of  flour  can  be  bought,  leaving  $7  unused  ;  or 
73067  barrels  can  be  bought  with  all  the  money. 

The  methods  of  proof  are  the  same  as  those  before  given. 

(b.)  When  the  above  explanation  and  the  principles  in- 
volved are  fully  understood,  the  names  of  the  denominations 
may  be  omitted  in  writing  the  work.     Thus  :  — 

8  is  contained  7  times  in  58,  with  2  remainder. 

8  is  contained  3  times  in  24,  with  no  remainder. 

8  is  contained  0  times  in  5,  with  5  remainder. 

8  is  contained  6  times  in  55,  with  7  remainder. 

Hence,  the  quotient  is  7306,  and  the  remainder  is  7  ;  or  the  complete 
quotient  is  73067. 

(c.)  In  the  following  form,  only  the  figures  of  the  quotient 
are  named. 

7  thousands,  3  hundreds,  0  tens,  6  units,  and  7  remaining ;  giving 
same  result  as  before. 

2.    If  8  acres  of  land  cost  $9707,  what  will  1  acre  cost  ? 

Reasoning  Process.  —  If  8  acres  of  land  cost  $9707,  1  acre  will  cost 
^  of  $9707. 

This  may  be  found  by  dividing  by  8,  as  in  the  last  example ;  or  it 
may  be  found  by  the  following  method.* 


*  The  student  should  observe   that  the  real  work  performed,  and  th* 
figures  used  in  writing  it,  are  the  same  by  one  method  as  by  the  other. 


104  DiyisiON. 

WRITTEN   WORK. 

Divisor    8  )  $9707       Dividend  =  cost  of  8  acres. 
$1213f     Quotient  =  cost  of  1  acre. 

We  divide  thus  :  ^  of  9  thousands  =  1  thousand,  with  a  remainder 
of  1  thousand.  1  thousand  =  10  hundreds,  and  7  hundreds  added  =  17 
hundreds.  ^  of  17  hundreds  =  2  hundreds,  with  a  remainder  of  1  hun- 
dred. 1  hundred  =  10  tens,  and  there  are  no  tens  to  add.  ±  of  10 
tens  =  1  ten,  with  a  remainder  of  2  tens.  2  tens  =  20  units,  and  7 
units  added  =  27  units.  ^  of  27  units  =  3  units,  with  a  remainder  of 
3  units. 

Therefore,  the  quotient  is  $1213,  and  there  is  a  remainder  of  $3  ;  or, 
since  3  -!-  8  =  f ,  we  may  say  that  the  quotient  is  $1213f . ,  Hence,  1 
acre  will  cost  $1213  3,  when  8  acres  cost  $9707. 

Note.  —  It  is  obvious  that  in  getting  1  of  the  above  number,  we 
have  really  divided  it  by  8. 

(d.)  By  omitting  the  names  of  the  denominations,  we 
should  have  — 

^  of  9  =  1,  with  1  remaining. 

|.  of  17  =  2,  with  1  remaining. 

i  of  10=1,  with  2  remaining. 

1  of  27  =  3,  with  3  remaining. 

This  gives  1213,  with  3  remaining,  or  1213^,  as  before. 

(e.)    By  only  naming  the  quotient  figures,  we  have  — 
1   thousand,  2  hundreds,  1  ten,  3  units,  and  3  remainder;   which 
gives  the  same  answer  as  before. 

Proof.  —  If  1  acre  costs  $1213|,  8  acres  will  cost  8  times  $12133  ;  or, 
which  is  the  same  thing,  8  times  $1213,  plus  $3.  This  will  be  $9707, 
if  the  work  is  correct. 

87,    Decimal  Fractions  in  the  Quotient. 

(a.)  The  answer  to  each  of  the  above  questions  might 
have  been  obtained  in  another  form,  by  reducing  the  final 
remainder  to  lower  denominations. 

{b.)  In  the  first  example,  the  7  units  remaining  =  70  tenths,  and  8  is 
contained  in  70  tenths  8  tenths  times,  with  a  remainder  of  6  tenths. 
But  6  tenths  =  60  hundredths,  and  8  is  contained  in  60  hundredths  7 
hundredths  times,  with  a  remainder  of  4  hundredths    But  4  hundredthi 


DIVISION.  105 

»e=  40  thousandths,  and  8  is  contained  in  40  thousandths  5  thousandths 
times.     This  V70uld  give  the  following  written  work  :  — 
Divisor    S8  )  $58455.000  =  Dividend. 

7306.875  =  Quotient  =  number  of  barrels. 

\c.)  In  the  second  example,  the  3  units  remaining  equal  30  tenths, 
and  1  of  30  tenths  =  3  tenths,  with  a  remainder  of  6  tenths.  But  6 
tenths  =  60  hundredths,  and  I  of  60  hundredths  =  7  hundredths,  with 
a  remainder  of  4  hundredths.  But  4  hundredths  =  40  thousandths,  and 
^  of  40  thousandths  =  5  thousandths.  This  would  give  the  following 
written  work :  — 

Divisor  =  8  )  $9707.000  =  Dividend. 

$1213.375  =  Quotient  =  cost  of  1  acre. 

(d.)  The  answers  in  this  form  can  be  proved  in  precisely 
the  same  way  as  were  the  former  answers. 


88.    Compound  Division. 

(a.)  All  questions  in  compound  division  must  of  necessity 
belong  to  the  second  class. 

A  benevolent  society  divided  10  cwt.  3  qr.  20  lb.  of  flour 
equally  among  6  poor  persons.  What  was  each  person's 
share  ? 

Reasoning  Process.  —  If  6  persons  had  10  cwt.  3qr.  201b.  of  flour,  1 
pereon  would  have  l  of  10  cwt.  3  qr.  20  lb. 

WRITTEN   WORK, 
cwt.    qr.    lb. 
6  )  10     3     20        —  share  of  6  persons. 

— oz. 

13       7     8  =  share  of  1  person. 

Explanation  of  Process,  i  of  10  cwt.  =  1  cwt.,  with  a  remainder  of 
4  cwt.    But  4  cwt.  =  16  qr.,  and  3  qr.  added  =  19qr.     ^  of  19qr.  = 

3  qr..  with  a  remainder  of  1  qr.  But  1  qr.  =  25  lb.,  and  20  lb.  added  = 
45  lb. ;  1  of  45  lb.  =  7  lb.,  with  a  remainder  of  3  lb.  But  3  lb.  =  48  oz., 
and  1  of  48  oz.  =  8  oz. 

Hence,  the  quotient  is  1  cwt.  3  qr.  7  lb.  8  oz. 

Second  Explanation.      6  is   contained  in  10  once,  with  4  remainder. 

4  cwt.  =  16  qr.,  and  3qr.  added  =  19qr.  6  is  contained  3  times  in  19, 
with  1  remainder.  1  qr.  =  25  lb.,  and  20  lb.  added  =  45  lb.  6  is  con- 
tained  7  times  in  45,  with  3  remainder.    3  lb.  =  48  oz.,  and  6  is  cott« 


106  DIVISION. 

tained  8  times  in  48.    Hence,  the  quotient  is  1  cwt.  3  qr.  7  lb.  8  oz.,  as 
before. 

(5.)  This  may  t3  proved  as  were  the  examples  in  simple 
numbers. 

Note.  —  Had  the  division  been  carried  no  farther  than  to  pounds, 
the  answer  would  have  been  1  cwt.  3  qr.  7  lb.,  with  a  remainder  of  3  Ib.^ 
or  1  cwt.  3  qr.  7  3.  lb. 

80.    Problems  for  Solution,  including  Reduction  Ascending, 

What  is  the  quotient  — 


1. 

Of  87543  -^  7  ? 

6. 

Of  348724-^4? 

2. 

Of  59487  ~  3  ? 

7. 

Of  37927-^6? 

3. 

Of  38640 -^  6? 

8. 

Of  29684  -^  9  ? 

4. 

Of  827546  —  9? 

9. 

Of  20034  -r-  6  ? 

5. 

Of  217483  H-  8  ? 

10 

Of  7248396275436002( 

)31 

-1- 

9? 

11.  How  many  yards   of   German  broadcloth  at  $6  per 
yard  can  be  bought  for  $174564  ? 

12.  How  many  sheep  at   $7  apiece   can   be   bought  for 
$7833  ? 

13.  If  8  melons   cost  $1,  how  many  dollars  will   3744 
melons  cost  ? 

14.  When  6  books  are  bought  for  $1,  how  many  dollars 
must  be  paid  for  1743  books  ? 

15.  How  many  kegs,  each  holding  9  gallons,  can  be  filled 
from  4752  gallons  of  molasses  ? 

16.  How  many  hours  will  it  take  a  vessel  to  sail  3937 
miles,  if  she  sails  8  miles  per  hour  ? 

17.  How  many  clocks  at  $6  apiece  can  be  bought  for 
$3528? 

18.  How  many  bags  of  coffee  at  $9  per  bag  can  be  bought 
for  $7487  ? 

19.  How  many  casks  of  raisins  at  $7  per  cask  can  be 
bought  for  $6594  ? 

20.  How  many  hours  would  it  take  a  man  who  walks  at 
•the  rate  of  4  miles  per  hour  to  walk  4739  miles  ? 


DIVISION.  107 

21.  How  many  hats  at  $4  apiece  can  be  bought  for 
$1372  ?     . 

22.  31741  quarts  =  how  many  bushels,  pecks,  and  quarts? 

WBITTEN  WORK. 

8)31741  — 5  qt 

4  )    3967  —  3  pk. 

991  bu. 

Reasoning  Process  and  partial  Explanation.  —  Since   8  qt.  =  1  pk., 

31741  qt.  must  equal  as  many  pecks  as  there  are  times  8  in  31741,  wliich, 

found  by  the  usual  method,  gives  3967  pk.  5  qt.    We  now  reduce  the 

pecks  to  bushels.     Since  4  pk.  =  1  bu.,  3967  pk.  must  equal  as  many 

bushels  as  there  are  times  4  in  3967,  which,  found  by  the  usual  method, 

gives  991  bu.  3  pk.    Hence,  31741  qt.  =  991  bu.  3  pk.  5  qt. 

Note.  —  Problems  like  the  above,  in  which  it  is  required  to  find  the 
value  of  a  number  of  units  of  one  denomination  in  terms  of  some 
higher,  are  ordinarily  called  Problems  in  Reduction  Ascending ;  but  they 
do  not  differ  in  their  nature  from  other  problems  in  division. 

23.  87633  days  =  how  many  weeks  ? 

24.  698472  inches  =  how  many  feet? 

25.  98464  5  =  how  many  ounces  ? 

26.  7987536  square  feet  =  how  many  square  yards? 

27.  97875  quarters  =  how  many  yards  and  quarters  r 

28.  87637  quarts  =  how  many  bushels,  pecks,  and  quarts? 

29.  798953  gills  =  how  many  gal.,  qt.,  pt.,  and  gi.  ? 

30.  793527  9  =z  how  many  tb,  3 ,  'S,  and  9  ? 

31.  587537  qr.  =  how  many  £,  s.,  d.,  and  qr.? 

32.  23975  fur.  r=  how  many  le.,  m.,  and  fur.? 

33.  15951  na.  ^  how  many  yd.,  qr.,  and  na.  ? 

34.  63359  pt.  =:  how  many  bu.,  pk.,  qt.,  and  pt.  ? 

35.  57984  gi.  m  how  many  gal.,  qt.,  pt.,  and  gi.  ? 

36.  7951  qr.  =  how  many  £.,  s.,  d.,  and  qr.  ? 

37.  4375  9  z=  how  many  lb,  § ,  5,  and  9  ? 

38.  A  farmer  put  75  gal.  3  qt.  1  pt.  2  gi.  of  cider  into 
bottles,  each  containing  1  pt.  2  gi.  How  many  bottles  did  it 
take  to  contain  it  ? 

Reasoning  Process.  —  Since  it  takes  1  bottle  to  hold  1  pt.  2  gi.,  it  must 
take  as  many  bottles  to  hold  75  gal.  3qt.  1  pt.  2gi.  as  there  are  times 
1  pt.  2  gi.  in  75  gal.  3  qt.  1  pt.  2  gi. 


108  DIVISION. 

Note.  —  Before  performing  the  division,  each  quantity  mnst  be 
reduced  to  gills. 

39.  A  grain  dealer  put  498  bu.  3  pk.  of  grain  into  bags, 
each  holding  1  bu.  3  pk.     How  many  bags  did  he  fill  ? 

40.  I  bought  536  yd.  2  qr.  2  na.  of  tape,  which  I  cut  into 
pieces  2  qr.  1  na.  long.     How  many  pieces  did  it  make  ? 

41.  If  9  men  earn  $4375.26  in  a  year,  how  much  will  1 
man  earn  in  the  same  time  ? 

Reasoning  Process.  —  If  9  men  earn  $4375.26  in  a  year,  1  man  will  earn 
1  of  $4375.26  in  the  same  time,  which  may  be  found  by  dividing  by  9. 

42.  If  7  cases  of  cloth  cost  $6545,  how  many  dollars  will 
1  case  cost  ? 

43.  If  185791b.  of  hay  are  obtained  from  9  acres,  how 
many  pounds  are  obtained  from  1  acre  ? 

44.  If  7  horses  can  draw  16786  lb.,  how  many  pounds  can 
1  horse  draw  ? 

45.  If  it  cost  $173549  to  build  9  miles  of  railroad,  how 
much  will  it  cost  to  build  1  mile  ? 

46.  A  father  divided  his  estate,  valued  at  $879643,  equally 
among  his  5  children.     What  was  the  share  of  each  ? 

47.  What  is  f  of  975646  ? 

48.  What  is  i  of  39547  bushels  ? 

49.  What  is  ^  of  7354.278  tons  ? 

50.  What  is  ^  of  12456.78  miles  ? 

61.   If  6  piano-fortes  cost  $1837.44,  how  much  will  7  cost  ? 

Reasoning  Process.  —  If  6  piano-fortes  cost  $1837.44,  1  will  ccst  | 
of  $1837.44,  which  (found  by  dividing  by  6)  is  $306.24.  If  1  piano-forte 
costs  $306.24,  7  will  cost  7  times  $306.24,  which  (found  by  multiplying 
by  7)  is  $2143.68. 

52.  If  9  wagons  cost  $1378.71,  what  will  4  cost? 

53.  If  8  gold  watches  cost  $775.36,  what  will  9  cost? 

54.  If  1  acre  of  land  costs  $327.52,  what  will  3  roods 
cost  ? 

55.  What  is  the  cost  of  5  cord  feet  of  wood  at  $7,376  per 
cord? 


DIVISION.  100 

56.  How  far  will  a  man  travel  in  6  days,  if  he  travels  at 
the  rate  of  174.79  miles  per  week? 

57.  If  1  bushel  contains  2150.4  cubic  inches,  how  many 
cubic  inches  will  3  pecks  contain  ? 

58.  If  1  peck  contains  537.6  cubic  inches,  how  many  cubic 
inches  will  a  six-quart  basket  contain  ? 

59.  If  in  1  quart,  liquid  measure,  there  are  57.75  cubic 
inches,  how  many  cubic  inches  are  there  in  1  qt.  1  gi.,  or  9 
gills? 

60.  If  1  man  can  perform  a  piece  of  work  in  2944  hours, 
in  how  many  hours  can  8  men  perform  the  same  work  ? 

Note.  —  It  is  obrious  that  8  men  will  perform  it  in  i  of  the  number 
of  hours  required  by  1  man. 

61.  If  it  takes  1  laborer  3474  days  to  earn  $5211,  how 
many  days  will  it  take  6  laborers  to  earn  the  same  sum  ? 

62.  In  a  certain  fort  there  are  provisions  enough  to  keep 
1  company  of  76  soldiers  7256  days.  How  long  can  8  com- 
panies of  the  same  size  be  kept  on  them  ? 

63.  10  men  bought  a  stack  of  hay  weighing  4  T.  16  cwt. 
0  qr.  1  lb.  14  oz.  6  dr.  What  ought  each  man's  share  to 
weigh  ? 

64.  A  farmer  had  9  equal  bins  filled  with  grain.  They  all 
contained  2047  cu.  ft.  1436  cu.  in.  How  many  feet  and  inches 
did  each  bin  contain  ? 

Q5.  The  brig  Maria  sailed  from  Philadelphia  to  Boston, 
with  207  T.  13  cwt.  2  qr.  19  lb.  of  coal,  |  of  which  was  land- 
ed at  one  wharf,  and  the  remainder  at  another.  What  was 
the  weight  of  that  landed  at  the  first  wharf?  at  the  second? 

66.  If  John  walks  ^  as  fast  as  George,  how  far  will  John 
walk  while  George  is  walking  97  m.  6  fur.  38  rd.  ? 

67.  A  company  of  9  California  gold  diggers  found  in  1 
month  371b.  4oz.  16dwt.  17  gr.  of  gold,  which  they  divided 
equally.     What  was  the  share  of  each  ? 

68.  8  men  bought  17  T.  18  cwt.  3  qr.  161b.  of  sugar, 
which  they  divided  equally.  What  was  the  share  of  each 
man  ? 

10 


110  DIVISION. 

69.  If  7  cows  eat  16  T.  5  cwt.  1  qr.  181b.  of  hay  in  1 
year,  how  much  will  9  cows  eat  in  the  same  time  ? 

70.  If  6  equal  pieces  of  cloth  contain  185  yd.  2  qr.  2  na,, 
how  much  will  5  pieces  of  the  same  size  contain  ? 

71.  If  5  pieces  of  Jbroadcloth  cost  £^7  13  s.  9  d.,  haw 
much  will  7  pieces  cost  ? 

72.  What  is  i  of  374  lb.  11  oz.  19  dwt.  20  gr.  ? 

73.  What  is  ^  of  857  m.  5  fur.  36  rd.  4  yd.  2  ft.  6  in.  ? 

74.  What  is  I  of  17  yr.  135  da.  14  h.  37  m.  56  sec.  ? 

75.  What  is  ^  of  15°  35' 54"? 

76.  What  is  i  of  191b  Hi  65  29  15 gr.? 

00.    Division  hy  Factors, 

(a.)  When  the  divisor  is  the  product  of  factors,  it  will 
sometimes  be  convenient  to  divide  by  its  factors,  instead  of 
dividing  by  the  whole  number  at  once. 

(6.)  To  understand  the  process  of  division,  and  method  of  getting 
the  true  remainder,  reduce  15262  quarts  to  bushels  and  quarts,  by  first 
reducing  it  to  bushels,  pecks,  and  quarts,  as  before  explained,  and  com- 
pare the  explanation  and  work  with  the  explanation  and  work  of  the 
following  example. 

(c.)    What  is  the  quotient  of  15262  divided  by  32  ? 

WRITTEN    WORK. 

8  )  15262  —  6,  the  first  remainder. 

4  )    1907  times  8,  —  3  times  8,  or  24,  the  2d  remainder. 

476  times  32  =  quotient  sought. 

24  -f-  6  =  30  =:  true  remainder. 

Explanation.  —  Since  32  =  4  times  8,  we  may  divide  by  8,  and  then 
by  4.  Dividing  by  8  gives  1907  for  a  quotient  and  6  for  a  remainder, 
i.  e.,  1907  times  8,  or  parts  of  8  each,  with  6  units  remaining  undivided. 

Dividing  1907  by  4  gives  476  for  a  quotient  and  3  for  a  remainder, 
i.  e.,  476  parts  each  equal  to  4  of  the  former  ones,  or  to  4  times  8,  or 
32,  with  3  times  8,  or  24,  undivided.  Adding  24  to  the  former  remain- 
der gives  30  for  the  true  remainder.  Hence,  15262  =  476  times  32, 
with  a  remainder  of  30,  or  it  equals  4762.2.  times  32. 

(rf.)  Had  the  problem  been  to  find  ^  of  15262,  the  first  quotient 
would  have  r^resentcd  the  number  of  units  in  each  part  obtained  hy 


DIVISION.  Ill 

dividing  15262  into  8  equal  parts;  and  the  second  the  number  of  units 
in  each  part  obtained  by  dividing  each  of  these  8  parts  into  4  others, 
or,  which  is  the  same  thing,  the  number  of  units  there  would  be  in  each 
part  obtained  by  dividing  15262  into  32  equal  parts. 

To  get  the  true  remainder,  observe  that  as  each  of  the  8  parts  givea 
a  remainder  of  3  units,  all  of  them  must  give  a  remainder  of  8  times  3, 
or  24,  which,  added  to  the  6  left  by  first  division,  gives  30  for  the  truo 
remainder. 

The  work  would  be  written  thus  :  — 
8  )  15262  —  6  first  remainder. 

4  )     1907  units  in  each  of  8  parts  —  3  remainder  on 

[each  part,  or  8  times  3  in  all. 

476  units  in  each  of  32  parts. 
24  -f-  6  =  true  remainder, 
(c.)  From  the  above,  it  is  evident  that  the  true  quotient 
will  be  obtained  by  dividing  the  dividend  by  one  factor,  the 
quotient  of  this  division  by  another,  the  quotient  of  the  last 
division  by  another,  and  so  on,  till  all  the  factors  of  the  set 
considered  are  used. 

(y*.)  The  true  remainder  may  be  obtained  by  multiplying 
the  remainder  of  each  division  by  the  divisors  of  all  the  pre- 
ceding divisions,  and  adding  the  products  to  the  remainder  of 
the  first  division. 


{g.)    What  is  the  value  — 

1.    Of  74385  -^4^? 

6. 

Of  57482  H-  15  ? 

2.   Of  67849  -^  35  ? 

7. 

Of  28654  -^  21  ? 

3.    Of  4535  -^  24  ? 

8. 

Of  5477  ~-  18  ? 

4.   Of  ^V  of  59358? 

9. 

Of  ^V  of  20249  ? 

5.   Of  2V  of  39475  ? 

10. 

Of  iV  of  172783 

f 


(k.)  The  most  important  application  of  the  division  by 
factors  is  made  when  the  divisor  is  a  multiple  of  10,  or  of 
some  power  of  10.  In  such  cases  we  first  divide  by  the 
power  of  10,  and  then  that  quotient  by  the  other  factor  of  the 
divisor,  getting  the  true  remainder  as  before  explained. 

11.    What  is  the  quotient  of  578635  4-  800  ? 

Explanation. — Dividing  by  100,  by  removing  the  point  two  places  to 
the  left,  gives  5786  for  a  quotient  and  35  for  a  remainder,  and  dividing 


112  DIVISION. 

this  quotient  by  8.  gives  a  final  quotient  of  723  and  a  remainder  of  2 
times  100,  or  200,  which,  added  to  35,  the  first  remainder,  gives  235  a« 
the  true  remainder. 

The  work  may  be  written  thus ;  — 

100  )  578635  —  35  "^ 

'       2  X  100  +  35  =  235. 

8  )  5786  "    '  ^ 


}- 


723 

(t.)    What  is  the  quotient  — 
12.   Of  6379  -^-40? 
18.   Of  27476  -^  300  ? 

14.  Of  427875-4-  9000? 

15.  Of  258647  -4-  80000  ? 


16.  Of  174326  -^  50  ? 

17.  Of  927673  -^  7000  ? 

18.  Of  42700-4-  9000? 

19.  Of  23750000-4-30000? 


91,    Divisor  a  large  Number. —  Trial  Divisor. 

(a.)  When  the  divisor  is  a  large  number,  it  is  not  always 
easy  to  tell  at  once  what  is  the  true  quotient  figure.  In  such 
cases,  the  number  expressed  by  one  or  two  of  the  left  hand 
figures  of  the  divisor  may  be  selected  as  a  sort  of  trial  divi' 
sor ;  but  to  determine  whether  the  quotient  figure  thus  ob- 
tained is  the  true  one,  it  will  be  necessary  to  find  the  product 
ifc^  of  the  divisor  by  the  quotient,  and  the  remainder  after  sub- 
tracting this  product  from  the  dividend. 

(6.)  The  product  of  the  divisor  by  the  quotient  should 
either  be  equal  to,  or  less  than,  the  dividend.  When  it  is 
equal  to  the  dividend,  the  division  can  be  exactly  performed ; 
but  when  it  is  less,  there  is  a  remainder,  which  may  be  found 
by  subtracting  it  from  the  dividend. 

(c.)  If,  in  any  case,  the  product  of  the  drvisoi*  by  the  sup- 
posed quotient  is  greater  than  the  dividend,  it  shows  that  the 
divisor  is  not  contained  as  many  times  in  the  dividend  as  the 
supposed  quotient  indicates. 

(c?.)    We  will  now  apply  these  principles  to  a  few  examples. 

1.    What  is  the  quotient  of  387  -4-  43  ? 

Solution.  —  Since  43  differs  but  little  from  4  tens,  we  may  infer  that 
the  entire  part  of  the  quotient  of  387  -t-  43  will  be  nearly  or  precisely 


DIVISION.  113 

the  same  as  that  of  38  -4-  4,  which  is  9.  To  ascertain  whether  this  be 
correct,  we  multiply  43  by  it,  which  gives  exactly  387,  and  proves  our 
work. 

The  figures  may  be  written  thus  :  — 
Divid. 
Divisor  =  43  )  387  (  9  =  Quotient 
387  =  9  times  43. 

2.  What  is  the  quotient  of  6169  -^  825  ? 

Solution. —  Since  825  differs  but  little  from  8  hundreds,  we  may  infer 
that  the  entire  part  of  the  quotient  of  6169  -f-  82.5  will  be  nearly  or  pre- 
cisely the  same  as  that  of  61  -i-  8,  which  is  7.  To  ascertain  if  this  is 
correct,  we  multiply  825  by  7,  which  gives  5775  for  a  product.  This  be- 
ing less  then  the  dividend  shows  that  the  quotient  figure  is  not  too 
large.*  Subtracting  this  product  from  the  dividend  gives  394  for  a 
remainder,  which,  being  less  than  the  divisor,  shows  that  the  quotient 
figure  is  correctf 

Hence,  the  quotient  is  7,  and  the  remainder  is  394,  or  the  complete 
quotient  is  7^4. 

WRITTEN  WORK. 

Divid. 
Divisor  =  825  )  6169  (  7  =  Quotient. 
5775  =:  7  times  825. 

394  =  Remainder. 

3.  What  is  the  quotient  of  55673  -^  6349  ? 

Solution. —  Making  6  the  trial  divisor,  we  have  9  for  a  trial  quotient. 
Multiplying  6349  by  9  gives  57141  for  a  product,  which,  being  greater 
than  the  dividend,  shows  that  the  quotient  figure  is  too  large.  Select- 
ing 8  as  the  quotient  figure,  and  multiplying  as  before,  gives  50792  for 
a  product,  which,  being  less  than  the  dividend,  shows  that  8  is  not  too 
large.     Subtracting  50792  from  55673  gives  a  remainder  of  4881. 

Hence  the  quotient  is  8,  and  the  remainder  is  4881,  or  the  complete 
quotient  is  8|8  8^. 

WRITTEN  WORK. 

Divid. 
Divisor  =  6349  )  55673  (  8  =  Quotient. 
50792  =  8  times  6349 

4881  =  Remainder. 


♦  For  it  shows  that  the  divisor  taken  7  times  is  less  than  the  dividend, 
t  For  it  shows  that  if  the  divisor  had  been  taken  once  more,  the  result 
would  have  been  more  than  6169. 
10   * 


114 


DIVISION. 


4.   What  is  tlie  quotient  of  3913  -^  482  ? 

Solution. —  Since  the  divisor  is  nearer  to  5  hundreds  than  4  hundreds, 
we  select  5  as  the  trial  divisor.  This  gives  7  for  a  trial  quotient,  and 
multiplying  gives  3374  for  a  product,  which,  being  less  than  the  divi- 
dend, shows  that  7  is  not  too  large.  Subtracting  3374  from  3913  gives 
a  remainder  of  539,  which,  being  greater  than  the  divisor,  shows  that  7 
is  too  small.  Substituting  8  in  its  place,  and  multiplying  and  subtract- 
ing as  usual,  gives  the  following  written  work  :  — 
482  )  3913  (8 

3856  =  8  X  482. 

57  =  Remainder. 
Hence,  the  quotient  is  8,  and  the  remainder  is  57,  or  the  complete 
quotient  is  8  5  7 


What  is  the  quotient  — 

5.  Of  5279  ~  847  ? 

6.  Of  19476-^2784? 

7.  Of  35947  -^8912? 
11.   What  is  the  quotient  of  15078  -^  276  ? 

WRITTEN    WORK. 


8.  Of  394687-^182578? 

9.  Of  807436-^294367? 
10.   Of  42974  -^  8523  ? 


276  )  15078  (  54 

1380    =  50  X  276 


1278 

1104  =  4  X  276 


174  ^  Remainder. 

Solution.  —  Making  3  the  trial  divisor,  we  may  infer  that  as  3  is  con- 
tained 5  times  in  15,  276  must  be  contained  5  tens  times  in  1507  tens. 
Writing  5  as  the  tens'  figure  of  the  quotient,  we  multiply  the  divisor  by 
it,  and  place  the  product,  which  is  1380,  under  the  tens  of  the  dividend, 
to  show  that  its  denomination  is  tens.  Subtracting  this  from  1507  tens 
leaves  a  remainder  of  127  tens,  to  which  adding  the  8  units  gives  1278 
units  to  be  divided. 

Since  3  is  contained  4  times  in  12,  we  may  infer  that  276  is  contained 
4  times  in  1278.  Writing  4  as  the  units'  figure  of  the  quotient,  we  mul- 
tiply the  divisor  by  it,  and  write  the  product,  which  is  1104,  under  the 
dividend.  Subtracting  leaves  a  remainder  of  174,  which  is  less  thau 
the  divisor. 

As  we  have  now  considered  all  the  denominations  of  the  dividend, 
we  may  consider  the  division  complete,  (unless  we  wish  tq  have  the 


DIVISION.  115 

quotient  appear  in  the  form  of  a  decimal  fraction.)    Hence,  the  quotient 
is  54,  and  the  remainder  is  174,  or  the  complete  quotient  is  541.X4. 

Note.  —  It  is  obvious  that  in  performing  the  above  work,  we  sub- 
tracted 50  times  the  divisor,  and  then  4  times  the  divisor,  which  is 
equivalent  to  54  times  the  divisor,  from  the  dividend,  and  that  this  left  a 
remainder  of  1 74. 

(e.)  Had  it  been  desirable  to  have  the  answer  contain  a 
decimal  fraction,  we  should  have  continued  the  division, thus:  — 

174  units  =  1740  tenths,  and  as  3  is  contained  5  times  in  17,  we  may 
infer  that  276  is  contained  5  tenths  times  in  1740  tenths.  Multiplying 
the  divisor  by  5,  and  subtracting  the  product  as  before,  gives  a  remain- 
der of  360,  which,  being  greater  than  the  divisor,  shows  that  the  divisor 
would  havCybeen  contained  more  than  5  tenths  times  in  the  dividend. 
"We  therefore  substitute  6  for  5  in  the  tenths  place  of  the  quotient,  and 
having  erased  the  last  product  and  remainder,  multiply  the  divisor  by  6, 
subtracting  the  product  as  before.  This  gives  a  remainder  of  84  tenths, 
which  may  be  reduced  to  hundredths,  and  divided  as  already  explained. 

In  the  following  form,  the  di\dsion  is  carried  out  to  millionths. 
276  )  15078.  (  54.630434 

1.380  =  5  tens  X  276  . 


1278. 
1104.  =  4 

times  276 
6  tenths  X  276 
:  3  hundredths  X  276 
=  4  ten-thousandths 

174.0 
165.6  = 

8.40 

8.28  = 

.1200 
.1104 

X  276 

960  =  hundred-thousandths 

828  =  3  hundred  thousandths  X  276. 

1320  =  millionths 

1104  =  1  millionth  X  276 


216  =  millionths  =  Remainder. 
Note.  —  It  is  obvious  that  we  have  subtracted  from  the  dividend  the 
iroducts  of  the  divisor  multiplied  by  5  tens,  by  4  units,  by  6  tenths,  by 
3  hundredths,  by  4  ten-thousandths,  by  3  hun  Ired-thousandths.  and  by  1 
millionth  ;  and  that  this  leaves  a  ren.aindei  of  44  millionths.  Hence, 
if  the  work  be  correct  the  sura  of  these  se  eral  products  added  to  the 
final  remainder  will  equal  the  dividend.  The  methods  of  proof  before 
explained  will  also  apply. 


116  DIVISION. 

O^,    Long  Division. 

[a.)  When,  as  in  85  to  01,  we  write  only  the  divisor, 
dividend,  quotient,  and  final  remainder,  the  process  is  called 
Short  Division  ;  but  when,  as  in  91,  we  write  the  divisor, 
dividend,  and  quotient,  and  also  the  products  of  the  divisor  by 
the  successive  figures  of  the  quotient,  together  with  the  re- 
'niainders  obtained  by  subtracting  these  several  products  from 
the  corresponding  denominations  of  the  dividend,  the  process 
is  called  Long  Division. 

(J),)  Long  Division  differs  from  Short  Division  in  these 
respects,  viz.:  First.  That  having  obtained  a  quotient  fig- 
ure, we  multiply  the  divisor  by  it,  writing  the  product  under 
the  part  of  the  dividend  considered. 

Second.  That  we  subtract  this  product  from  the  part  of  the 
dividend  considered,  writing  the  remainder. 

Third.  That  we  write  the  next  figure  of  the  dividend  after 
this  remainder,  to  form  the  next  partial  dividend. 

(c.)  The  real  difference  between  them  may  be  stated  thus  : 
In  long  division  more  of  the  work  is  written  than  in  short 
division. 

{d.)  Long  division  is  generally  employed  when  the  divisor 
is  a  large  number. 

(e.)  If  at  any  time  the  product  of  the  divisor  by  a  quotient 
figure  should  be  less  than  the  part  of  the  dividend  considered, 
it  would  show  that  the  quotient  figure  was  too  large.  (See 
3d  example,  91.) 

(/.)  If  the  remainder  obtained  by  subtracting  from  the 
dividend  the  product  of  the  divisor  by  any  quotient  figure 
should  be  greater  than  the  divisor,  it  would  show  that  the 
q[uotient  figure  was  too  small.     (See  4th  example,  91.) 

{g.)  Perform  the  following  examples  by  long  division. 
See  solution  to  11th  example  in  91.) 

What  is  the  quotient  — 

1.  Of  3784  -^  21  ? 

2.  Of  59378  -^  82  ? 

3.  Of  386495  ~  289  ? 


4.   Of  43f)25  -h  714? 


5.  Of  386427  -^  5287  ? 

6.  Of  2700684  ~  19743? 

7.  Of  438277  -7-34? 

8.  Of  6293876  h-  2874  ? 


DIVISION.  117 

9.  How  many  square  feet  are  equal  to  42912  square 
inches  ? 

10.  How  many  boxes,  each  containing  73  quarts,  can  be 
filled  from  658764  quarts  of  meal? 

11.  How  many  watches  at  $48  apiece  can  be  bought  for 
$3456  ? 

12.  If  a  boy  can  perform  37  examples  in  one  day,  how 
many  days  will  it  take  him  to  perform  8764  examples  ? 

13.  If  a  vessel  sails  147  miles  per  day,  how  many  days 
will  it  take  her  to  sail  3579  miles  ? 

14.  If  252  pints  of  cider  will  fill  a  barrel,  how  many  bar- 
rels can  be  filled  from  876438  pints  ? 

15.  How  many  weights,  each  weighing  5Q  lb.,  will  be  re- 
quired to  balance  7  loads  of  iron,  each  weighing  9472  lb.  ? 

16.  James  can  pump  37  quarts  of  water  per  minute,  and 
William  can  pump  43  quarts  of  water  per  minute.  How 
many  minutes  will  it  ^ake  William  to  pump  as  much  as  James 
can  pump  in  473  minutes  ? 

17.  What  will  1  acre  of  land  cost  if  237  acres  cost 
$236526? 

18.  If  687  bushels  of  potatoes  cost  $515.25,  what  will  236 
bushels  cost  ? 

19.  If  67  yd.  of  silk  cost  $75,375,  what  will  48  yd.  cost? 

20.  If  89  barrels  of  flour  contain  17444  pounds,  how 
many  pounds  wiU  258  barrels  contain  ? 

03.    Abbreviated  Process. 

(a.)  By  performing  the  subtraction  at  the  same  time  that 
we  do  the  multiplication,  much  of  the  written  work  can  be 
avoided. 

(5.)  The  subtraction  may  be  performed  in  accordance  with 
the  principles  explained  in  73.  The  following  example 
will  illustrate  it.  We  have,  however,  given  only  the  mechan- 
ical process,  leaving  it  with  the  pupil  to  explain  the  various 
steps. 

1.   What  is  the  quotient  of  1947.63  -^  396? 


118 


DIVISION. 


Process.  396  is  contained  4  times  in  1947.  4  times  6  =  24,  which 
subtracted  from  27  leaves  3  ;  this  we  write  in  the  units'  place.  4  timea 
9  =  36,  and  2  added  make  38,  which  subtracted  from  44  leaves  6.  4 
times  3  =  12,  and  4  added  make  16,  which  subtracted  from  19  leaves  3. 
The  first  remainder  is,  therefore,  363.  Reducing  this  to  tenths,  and 
adding  the  6  tenths,  we  have  3636  tenths,  which  divided  by  396  gives  9 
tenths  for  a  quotient.  We  multiply  and  subtract  as  before,  and  thus 
proceed  till  the  division  is  completed,  or  till  we  have  as  many  decimal 
places  in  the  quotient  as  we  wish. 

WRITTEN   WORK. 

396  )  1947.63  (4.9182  =  Quotient. 
363.6 
7.23 
3.270 
.1020 

.0228  =:  Remainder. 
(c.)    Perform  the  division  in  the  following  examples  in  the 
same  way :  — 

2.  What  is  the  quotient  of  78643  -h  47  ? 

3.  What  is  the  quotient  of  137648  -f  326  ? 

4.  What  is  ^V  of  65427  ? 

5.  What  is  -^^^  of  3865726  ? 

6.  What  is  the  value  of  764379  cubic  inches,  expressed  in 
cubic  feet  and  inches  ? 

7.  How  many  square  feet  and  inches  are  equal  to  954376 
square  inches  ? 

8.  If  the  wages  of  96  laborers  for  1  year  are  $26470.08, 
what  will  be  the  wages  of  1  laborer  for  the  same  time  ? 

9.  If  875  pairs  of  boots  cost  $2737.16,  how  much  will  1 
pair  cost  ? 

10.  How  many  pairs  of  shoes  at  $2.37  per  pair  can  be 
bought  for  $184.86  ? 

Note.  —  In  solving  such  examples  as  the  above,  both  divisor  and 
dividend  should  be  reduced  to  cents,  thus  :  if  for  237  cents  one  pair  of 
shoes  can  be  bought,  for  18486  cents  as  many  pairs  can  be  bought  aa 
there  are  times  237  in  18486,  which  may  be  found  by  methods  already 
explained. 


CONTRACTIONS  AND  MISCELLANEOUS  PROBLEMS.  119 

11.  How  many  pounds  of  tea  at  $.47  per  pound  can  be 
bought  for  $464.23  ? 

12.  How  many  bushels  of  wheat  at  $1.85  per  bushel  can 
be  bought  for  $287.35  ? 

13.  How  many  pounds  of  raisins  at  $.125  per  pound  can 
be  bought  for  $39,875  ? 

14.  How  many  days  must  a  man  labor  at  $1.75  per  day 
to  earn  $596.75  ? 

15.  I  gave  17  acres  of  land  worth  $46.98  per  acre  in  ex- 
change for  wild  land  at  $4.59  per  acre.  How  many  acres  of 
wild  land  did  I  receive  ? 

16.  1475869  cu.  in.  =  how  many  C,  Cd.  ft.,  cu.  ft.,  and 
cu.  in.  ? 

17.  8383794  oz.  =  how  many  T.,  cwt.,  qr.,  lb.,  oz.,  and  dr.  ? 

18.  67458  sq.  rd.  =  how  many  A.,  K.,  and  sq.  rd.  ? 

19.  57864  gr.  :=.  how  many  lb.,  oz.,  dwt.,  and  gr.  ? 

20.  457986  cu.  in.  =  how  many  C,  Cd.  ft.,  cu.  ft.,  and 
cu.  in.  ? 


SECTION    YIII 


CONTRACTIONS  AND  MISCELLANEOUS   PROB- 
LEMS. 

Introductory  Note.  —  It  is  frequently  the  case  that  by  carefully  exam- 
ining the  numbers  we  are  to  operate  upon,  we  can  discover  some  abbre- 
viated method  of  performing  the  work.  We  have  already  suggested 
several  such  methods,  and  it  is  the  design  of  this  section  to  suggeac 
others. 

d4:.    Multiplier  a  convenient  fractional  Part  of  10 j  100, 
1000,  ^c. 

When  the  multiplier  is  a  convenient  fractional  part  of  10, 
100,  1000,  or  a  unit  of  any  higher  denomination,  the  follow- 
ing principles  may  be  advantageously  applied. 


J  20     CONTRACTIONS    AND    MISCELLANEOUS   PBOBJLEMS. 

1.  How  many  are  25  times  679  ? 

Solution. —  Since  25  equals  one  fourth  of  100,  25  times  679  must 
aqual  one  fourth  of  100  times  679,  or  1.  of  67900,  which  equals  16975. 

2.  How  many  are  25  times  657  ? 

3.  How  many  are  50  times  657  ? 

4.  How  many  are  12^  times  834  ?  (12^  =  ^  of  100.) 

5.  How  many  are  16f  times  957  H  (16f  =  ^  of  100.) 

6.  How  many  are  33^  times  871  ?  (33^  =  ^  of  100.) 

7.  How  many  are  14f  times  249  ?  (14^  ==  i  »/  100.) 

8.  How  many  are  3^  times  249  ?  (3^  =z  ^  of  10.) 

9.  How  many  are  2^  times  822  ?  (2^  =  i  o/  10.) 

10.  How  many  are  6^  times  944?     (6^  =^ -^  of  100.) 

11.  How  many  are  333^  times  9478?  (333^  =  ^  of 
1000.) 

12.  How  many  are  125  times  8767  ?     (125  m  ^  o/  1000.) 

13.  How  many  are  250  times  6894  ?     (250  =  i  of  1000.) 

0»5.     One  Part  of  MidtipUer  a  Factor  of  another  Paii, 

When  one  part  of  the  multiplier  is  a  factor  of  another  part, 
much  labor  can  often  be  saved  by  applying  the  principles 
illustrated  in  the  following  examples  :  — 

1.   What  is  the  product  of  7678  multiplied  by  427  ? 

Solution.  —  By  examining  the  multiplier,  we  perceive  that  it  equalt 
420  -|-  7  =  42  tens  -j-  7  units,  and  that  42  tens  =  6  tens  or  60  times 
7.    Hence,  we  have  the  following  written  work. 

a  =        7678 
b==  427 


7  times  a  =  c  =       53746 
6  tens,  or  60  times  c  =  420  times  a  =  d  =  322476 


c  +  d  =  b  times  a  =  e  =;  3278506    =  An». 

2.   What  is  the  product  of  72144  times  874369  ? 

Explanation.  —  By  examining  the  multiplier,  we  perceive  that  it 
quals  72  thousands  -f-  144,  and  that  as  144  =  twice  72,  it  must  equal 
times  001  of  72  thousands.    Hence, 


CONTRACTIONS    AND    MISCELLANEOUS    PROBLEMS.    121 

a  =  874369 

b=  72144 


72000  X  a  =  c  =  62954568000 
2  times  .001  of  c  =  144  times  a  =  d  =      125909136 


c  +  d  =  b  times  a  =  e  =  63080477136 

3.   How  many  are  1875625125  times  97643721785  ? 

a=         97643721785 
b=  1875625125 


125  times  a  =  c  =       12205465223125 
5000  times  c  =  625000  times  a  =  d  =    610273261 15625 
5000  times  d=  1875000000  times  a  =e  =  183081978346875 


c  -f  d  -f  e  =  b  times  a  =  183143017878455848125 
How  many  are  — 

4.  14874  times  376437  ? 

5.  19899  times  879438  ? 

6.  17525  times  43678  ? 

7.  83415  times  476437  ? 

8.  43821973  times  976327864? 

9.  36324  times  54289732  ? 

10.  1998999  times  463829748  ? 

11.  1752512^  times  83743954? 

96.    To  divide  hy  99,  999,  ^c. 

{a.)  Since  10  =  9  +  1,  and  100  =  99  +  1,  and  1000 
=  999  +  1,  &c.,  it  follows  that  40  z=:  4  times  9  +  4 ;  that 
2800  =  28  X  99  +  28 ;  that  37900  =  379  X  99  +  379 ; 
that  786000  =  786  times  999  +  786;  &c.  On  this  princi- 
ple the  following  processes  are  based. 

1.  What  is  the  quotient  of  437  -f-  99  ? 

Solution.  437  =  400  +  37 ;  but  400  =  4  times  99  4-  4.  Hence, 
437  =  4  times  99  -f-  4  +  37  =  4  times  99  -f  41  =  4|^  times  99  =s 
Answer. 

2.  What  is  the  quotient  of  15378  -^  99  ? 

Solution.      15378  =  15300  +  78  =  153  times  99  +   153  +  78  = 
153  times  99  -|-  231.     But  231  =  200  -f  31  ==  2  times  99  -|-  2  -f  31 
11 


122     CONTRACTIONS    AND    MISCELLANEOUS    PROBLEMS. 

=  2  times  99  +  33-  Now,  by  adding  153  times  99  to  2  times  99,  we 
have  155  times  99.  Hence,  15378  =  155  times  99  -f-  33  =  155|| 
times  99. 

The  following  exhibits  a  convenient  form  for  writing  the  work :  — 

153  78 
2  31 


155  33  =  155  3 1 
The  full  explanation  is  the  same  as  that  already  given  The  neces- 
sary numerical  operations  are  as  follows :  Separating  the  hundred* 
from  the  tens  and  units  by  a  vertical  line,  we  add  the  number  at  the 
right  of  the  line  to  that  at  the  left.  This  gives  153  +  78  =  231  =  first 
remainder.  Writing  this  beneath  the  dividend,  and  adding  as  before, 
we  have  2  -|-  31  =  33,  which,  being  less  than  99,  is  the  final  renjainder. 
The  sum  of  the  numbers  at  the  left  of  the  line,  or  153  +  2  =  155,  the 
quotient. 

(h.)    In  a  similar  manner  we  can  divide  by  999,  9999,  &c. 

f ^ )  Let  the  pupil  find,  if  he  can,  the  application  of  a 
similar  principle  to  the  division  by  98,  97,  96,  &c. ;  also  to 
998,  997,  &c.,  and  afterwards  perform  the  following  exam- 
ples :  — 

What  is  the  quotient  — 

3.  Of  186738  -i-  99  ?       ' 

4.  Of  49763842  -^  999  ? 

5.  Of  763852748  -^  9999  ? 

6.  Of  9842987483 -r  99999? 

7.  Of  54783  -^  98  ? 

8.  Of  2987637  —  96? 

9.  Of  248763-^-997? 
10.   Of  69874325  -^  9998  ? 

07,    To  divide  by  any  convenient  fr actional  Part  of  10, 

100,  or  1000. 

(a.)  Since  100  -^  25  =r  4,  900  -f-  25  must  equal  9  times 
4;  1700  -^  25  must  equal  17  times  4;  49600  -^  25  must 
equal  49  G  times  4. 

(h,)  Since  100  -^  12^  =  8,  3800  -r-  12J  must  equal  38 
times  8 ;  49700  -f-  12^  must  equal  497  times  8,  &c 

(c.)    Since  1000  -^  125  =  8,  479000  -r  125  must  equal 


CONTRACTIONS  AND  MISCELLANEOUS  PROBLEMS.  123 

479  times  8;  3785000  -^  125  must  equal  3785  times  8, 
&c. 

(«?.)  These  principles  are  applied  in  the  following  pro- 
cesses. 

1.    What  is  the  quotient  of  9738  -f-  25  ? 

Solution.     9738,  9700  +  38.      But  9700  =  97  times  100  =  97  times 

.1.3  times  25.     Hence,  9738  -*- 


25  =  388  4-  1  J-|  =  389^3. 

. 

What  is  the  quotient  — 

2.   Of  86794-^25? 

6. 

Of  54737  -^  125? 

3.   Of  3475  -^  121  ? 

7. 

Of  98734  -^  33^  ? 

4.   Of  6950  -^  16f  ? 

8. 

Of  4765465  H-  333^  ? 

5.   Of  42725  ^  6^  ? 

9. 

Of  657847  ^250? 

08.    Miscellaneous  Problems. 

1.  A  man  bought  a  house  lot  and  garden  for  $1378.24; 
he  paid  $4796.87  for  building  a  house,  $1274.38  for  building 
a  stable  and  carriage  house,  $438.47  for  fencing  the  lot, 
$578.37  for  laying  out  the  garden  and  grounds,  $1287.63 
for  a  span  of  horses  and  a  carriage,  $1328.56  for  furnishing 
his  house,  and  then  had  $47289.43  left.  Ho^  much  money- 
did  he  have  at  first  ? 

2.  Mr.  French  bought  a  house  for  $6742.38,  and  after  pay- 
ing $138.47  for  having  it  painted,  and  $527.94  for  repairs, 
he  sold  it  for  $8472.     Did  he  gain  or  lose,  and  how  much  ? 

3.  Mr.  Hall  bought  437  cords  of  wood  at  $3  per  cord,  and 
sold  it  at  $4.75  per  cord.  What  did  he  gain  by  the  specula- 
tion ? 

4.  A  man  started  on  a  journey  of  1164  miles,  and  trav- 
elled 37  miles  per  day  for  the  first  21  days.  How  many  days 
would  it  take  him  to  finish  the  journey  if  he  should  travel  at 
the  rate  of  43  miles  per  day  ? 

5.  Bought  43  acres  of  land  at  $28.73  per  acre,  and  sold  it 
at  $37.73  per  acre.     How  much  did  I  gain  ? 

6.  I  bought  487  yards  of  cloth  at  $2  per  yard,  and  627 
yards  at  $4  per  yard ;  I  sold  the  whole  of  it  at  $4  per  yard. 
How  much  did  I  gain  ? 


124     CONTRACTIONS    AND    MISCKLLANEOUS    PROBLEMS. 

7.  A  trader  bought  528  barrels  of  flour  at  $7  per  barrel, 
875  barrels  at  $8  per  barrel,  and  497  barrels  at  $9  per  bar 
rel.  He  sold  the  whole  of  it  at  $8  per  barrel.  Did  he  gain 
or  lose,  and  how  much  ? 

8.  How  much  will  a  house  lot,  137  feet  long  and  89  fe^t 
wide,  cost,  at  2  cents  per  square  foot  ? 

9.  Which  is  the  larger  13  X  17  X  28  X  43,  or  28  X  13 
X  43  X  17,  and  how  much  ? 

10.  A  man  who  had  $4376  invested  it  in  flour  at  $8  per 
barrel.  He  sold  238  barrels  of  the  flour  at  $9  per  barrel, 
and  the  rest  for  enough  to  make  up  $5232.  For  how  much 
did  he  sell  the  last  lot  per  barrel  ? 

11.  A  man  bought  a  horse  for  $237;  he  kept  him  19 
weeks  at  an  expense  of  $2.50  per  week,  and  then  exchanged 
him  for  another  horse,  receiving  $48  for  their  difference  in 
value.  After  keeping  the  second  horse  4  weeks,  at  an  ex- 
pense of  $2.25  per  week,  he  sold  him  for  $225.  Now,  allow 
ing  that  the  use  of  each  horse  was  worth  $1.75  per  week,  did 
he  gain  or  lose  by  the  transaction,  and  how  much  ? 

12.  To  12  times  487  add  8  times  683,  multiply  the  sum  by 
7,  and  subtract  4279  from  the  product. 

13.  James  Smith  bought  of  John  Brown  7  hogsheads  of 
molasses,  each  containing  147  gallons,  at  $.27  per  gallon,  and 
gave  in  payment  $100  in  money,  and  the  rest  in  nails  at  5 
cents  per  pound.     How  many  pounds  of  nails  did  it  take  ? 

14.  How  many  cords  of  wood  at  $4  per  cord  can  be  bought 
for  82  barrels  of  apples  at  $2  per  barrel  ? 

15.  John  earns  $4  per  week,  and  William  earns  $7.  How 
many  weeks  will  it  take  John  to  earn  as  much  as  William  can 
earn  in  48  weeks  ? 

16.  Gave  3  hogsheads  of  oil,  each  containing  167  gallons, 
worth  $1.68  per  gallon,  and  $200  in  money,  for  4  acres  of 
land.  How  much  ought  the  land  to  be  worth  per  acre,  that 
I  may  neither  gain  nor  lose  ? 

17.  George  had  378  oranges,  which  sold  at  2  cents  apiece. 
With  the  money  thus  received  lie  bought  a  box  of  oranges, 
which  he  found  contained  427.    He  ate  9  of  these,  gave  away 


CONTRACTIONS  AND  MISCELLANEOUS  PROBLEMS.  125 

7,  had  4  stolen  from  him,  sold  294  at  3  cents  apiece,  and  the 
rest  at  2  cents  apiece.  Did  he  gain  or  lose  on  the  box,  and 
how  much  ? 

18.  How  many  times  will  a  carriage  wheel,  12  feet  6  inches 
in  circumference,  revolve  in  going  5  miles,  4  fur.,  18  rd.,  4  yd., 
1  ft.,  1  in.  ? 

19.  A  boy,  riding  with  his  father,  ascertained  that  the 
hinder  wheel  of  the  carriage,  which  was  10  feet  4  inches  in 
circumference,  revolved  1297  times  in  passing  from  one  vil- 
lage to  another.  How  far  apart  were  the  villages,  reckoning 
the  distance  in  miles,  furlongs,  rods,  &c.  ? 

20.  Mr.  Clarke's  house  lot  is  83  feet  wide,  and  contains 
8051  square  feet.     How  long  is  it  ? 

21.  Mr.  Angell  says  that  his  house  lot  is  97  feet  long,  but 
that  if  it  were  100  feet  long,  it  would  contain  168  square  feet 
more  than  it  now  does.  How  many  feet  does  it  now  con- 
tain ? 

22.  Multiply  837  by  5,  add  247  to  the  result,  divide  this 
by  4,  and  calHng  the  result  dollars,  find  how  many  yards  of 
cloth  at  $2  per  yard  you  could  buy  with  it. 

23.  By  buying  a  cargo  of  coal  at  $6  per  ton,  and  selling  it 
at  S8  per  ton,  I  gained  $198.     How  much  did  I  pay  for  it? 

24.  A  silversmith  bought  13  lb.,  4  oz.,  2  dwt.,  5  gr.  of  sil- 
ver, and  after  mixing  with  it  1  lb.,  5  oz.,  15  dwt.,  19  gr.  of 
alloy,  made  it  into  spoons,  each  weighing  1  oz.,  4  dwt.,  17  gr. 
How  many  spoons  did  he  make  ? 

25.  I  bought  the  wood  standing  on  9  acres  of  land,  paying 
for  it  at  the  rate  of  $67.25  per  acre.  I  paid  $.625  per  cord 
for  having  it  cut,  and  $.75  per  cord  to  have  it  carted  to  a  rail- 
road depot,  where  I  sold  it  for  $4.50  per  cord.  If  there  was 
an  average  of  30  cords  to  the  acre,  how  much  did  I  gain  by 
the  transaction  ? 

26.  A  trader  mixed  536  lb.  of  sugar,  worth  9  cents  per 
pound,  with  52  lb.,  worth  6  cents  per  pound,  and  432  lb., 
worth  7  cents  per  pound.  For  how  much  per  pound  ought 
he  to  sell  the  mixture  so  as  neither  to  gain  nor  lose  ? 

11* 


126  BILLS    OF    GOODS. 

27.  A  trader  bought  597  gallons  of  vinegar  at  14  cents  pep 
gallon,  and  after  mixing  with  it  24  gallons  of  water,  sold  it 
for  15  cents  per  gallon.  How  much  did  he  gain  by  the  trans- 
actfon  ? 

28.  I  bought  a  pile  of  wood,  144  feet  long,  4  feet  wide, 
and  6  feet  high,  at  $3.92  per  cord.     What  did  it  cost  me  ? 

29.  How  many  square  feet  in  the  walls  of  a  room  18  feet 
long,  16  feet  wide,  and  12  feet  high  ? 

30.  If  William  walks  at  the  rate  of  16  rods  per  minute, 
and  Joseph  at  the  rate  of  19  rods  per  minute,  how  long  will 
it  take  Joseph  to  overtake  William,  when  William  has  12 
minutes  the  start. 


99.    Bills  of  Goods. 

{a.)  When  a  man  sells  goods,  he  usually  gives  the  pur- 
chaser a  written  statement  of  the  articles  bought,  and  the 
prices  he  is  to  pay  for  them.  Such  a  statement  is  called  a 
«  Bill  of  the  Goods,"  or  simply  a  "  Bill." 

{h.)  A  bill,  like  every  other  business  paper,  should  contain 
the  date,  i.  e.,  the  time  and  place  of  the  transaction,  and  also 
the  names  of  the  parties,  and  an  account  of  the  transaction. 
K  the  goods  are  paid  for,  the  bill  should  be  receipted ;  i.  e., 
the  words  Received  Payment  being  written  at  the  bottom,  the 
seller  should  affix  his  name. 

(c.)   The  following  example  will  illustrate  this  :  — 

Mr.  George  W.  Dodge  is  a  trader,  residing  in  Lancaster. 
On  the  1st  of  January,  1855,  he  sold  to  Mr.  Humphrey  Bar- 
rett, for  cash,  3  yards  of  broadcloth  at  $3.75  per  yard,  6  yards 
of  doeskin  at  $1.87  per  yard,  32  yards  of  sheeting  at  11  cents 
per  yard,  9  yards  of  black  silk  at  97  cents  per  yard,  and  6 
linen  handkerchiefs  at  34  cents  apiece. 

Mr.  Dodge  made  out  the  l)ill  as  follows :  — 


BILLS    OF    »OODS. 


127 


.tLo/ncaafet,     Zzn 

.  /,  /855 

K/^t.  i^bamAntev  S^aizetf, 

^ou^di  0/ 

'^.W. 

QJod^i. 

3     yc^a.   ^wadcUim     at    &3  //S 

///.i'J 

6        "       (^oe^d^          " 

/.87 

//.i'i' 

3S  "    S^La,:^ 

.// 

3.5S 

f        "       ^/acd  S'Ji  " 

■97 

B.73 

6       "       ^cnenJ^cM^" 

.34 

S.OA 

036.76 


(d,)  If  Mr.  F.  W.  Spofford,  Mr.  Dodge's  clerk,  had  re- 
ceived the  money  due  on  the  bill,  he  would  have  receipted  it 
by  some  form  similar  to  the  following  :  — 


Or, 


128  BILLS    OP    GOODS. 

(e.)  If  the  goods  had  been  bought  on  credit,  the  words 
"  Received  Payment"  might  have  been  written,  but  no  name 
would  have  been  affixed.  It  is  the  signing  of  the  receipt  by 
the  seller,  or  his  authorized  agent,  which  gives  validity  to  it. 
The  receipted  bill  should  be  kept  by  the  person  who  pays  it, 
as  evidence  of  the  payment. 

(/.)  The  following  form  of  heading  a  bill  is  often  used 
instead  of  the  above  :  — 

^anca^^z,    ^an,   /^   ^  855 . 

Note.  —  A  person  is  my  dd)tor  when  he  owes  me  money,  and  my 
creditor  when  I  owe  him. 

(g.)  When  articles  are  bought  or  services  rendered  at  dif- 
ferent times,  the  bill  may  be  headed  and  receipted  as  before, 
and  the  dates  of 'the  various  transactions  written  at  the  left, 
opposite  the  entries. 

(Ji.)  Theodore  Gay  is  a  trader  living  in  Dedham.  He 
sold  to  Samuel  French  the  following  articles,  viz. :  Jan.  3, 
1855,  5  bags  of  meal  at  $1.58  per  bag,  and  2  bags  of  corn  at 
$1.54  per  bag;  Jan.  27,  8  lbs.  of  coffee  at  16  cents  per 
pound;  Feb.  7,  3  lbs.  tea  at  59  cents  per  pound;  Feb.  21, 
1  bbl.  of  flour  for  $10.37,  and  14  lbs.  of  brown  sugar  at 
9  cents  per  pound.  On  the  1st  of  March,  Mr.  Gay  being  in 
want  of  money,  made  out  a  bill,  and  sent  it  to  Mr.  French, 
who  paid  it  on  the  3d  of  March. 

(i.)    The  bill  was  made  out  in  the  following  form  :  — 


BILL8    OF    GOODS.  129 

Uo  Uneoaoie   ^iS^Uj  ^^^. 

/855. 

Jan.  3,  ^o^S^cya  ^ea/,at^^.5S,  ^7-90 


C(          (( 

(f 

S      "      '^otn,    " 

^.5A, 

3.0'8 

S7, 

(( 

B/A.    ''Soffee,     " 

J6, 

^.SB 

Ml  7, 

cc 

3 /A.   ^ea, 

■59, 

/./; 

"  s/, 

C( 

////c%^,       " 

^0.37 

(t    (( 

(( 

.09, 

^.S6 

'aid  3,  ^855. 

p^  (y.)  If  Mr.  French  had  paid  Mr.  Gay  $2  on  the  1st  of 
February,  and  worked  for  him  the  4  days  ending  Feb.  24 
at  $1.50  per  day,  the  first  part  of  the  bill  would  have  been 
raade  out  as  before,  and  then  the  credit  entries  would  have 
been  made  as  follows  :  — 


130  BILLS    OK    GOODS. 

'^J.  i"/,  ^ot  ^A.  /^.  ^/'yat,  ad  ^.0^,  / "  /.S6 

.%/  /,  ^y   ''Sa.i,        .        .        .        .     /    2.00 
"  SA,     "      AciayJ  L^ot,at0/.5O,0    6.00 

/    B.OO 

Uneoaoze   ^^^y« 

(  k.)  The  mere  sending  of  a  bill  to  a  person,  except  at  his  request, 
or  at  the  time  of  sending  the  articles  for  which  the  biil  is  made  out,  la 
equivalent  to  a  request  that  the  money  due  on  it  should  be  paid. 

(Z.)  A  bill  is  said  to  be  against  the  person  who  owes,  and  in  favor  of 
the  one  who  is  to  receive  the  money  due  on  it.  Thus  the  first  of  the 
preceding  bills  is  against  Humphrey  Barrett,  and  ih  favor  of  Geo.  W. 
Dodge, 

100.    Examples  for  Practice.  —  Due  Bills. 
Make  out  the  proper  bills  for  each  of  the  following  exam- 
ples :  — 

1.  L.  H.  Holmes  is  a  dry  goods  dealer,  residing  in  Bridge- 
water.  Oct.  7,  1854,  he  sold  to  E.  C.  Hewett,  for  cash,  3 
yds.  of  broadcloth  at  $3.62,  3  yds.  of  doeskin  at  $1.68,  1 
cravat  for  $1.50,  1  vest  for  $6.00,  1  pair  of  gloves  for  $1.00, 
and  1  pair  of  boots  for  $4.50. 

2.  A.  B.  Curry  &  Son,  of  Providence,  sold  to  Geo.  A. 
Richards,  June  13,    1855,   the   following   articles,  viz.:   85 

♦  This  entry  is  repeated  so  as  to  make  the  form  of  the  bill  more  ap- 
parent. 


BILLS    OF    GOODS.  131 

lbs.  live  geese  feathers,  at  50  cents  per  pound;  12  common 
chairs,  at  $.42  each ;  6  cane  seat,  at  $1.00  each ;  6  mahogany 
spring  seat,  at  $3.00  each ;  3  common  bedsteads,  at  $3.00  each; 
2  cottage  bedsteads,  at  $5.00  each;  2  best  hair  mattresses,  25 
lb.  each,  at  $.50  per  lb. ;  2  palm  leaf  mattresses  at  $4  each,  2 
husk  at  $5.00  each,  2  straw  at  $2.50  each,  1  sofa,  $35 ;  1 
pair  tete-a-tetes,  $75 ;  1  gilt  mirror,  $75  ;  1  marble  centre 
table,  $42  ;  1  secretary,  $45  ;  1  painted  chamber  set,  $45  ; 
1  enamelled  chamber  set,  $125.00;  3  common  bureaus  at 
$8.00  each ;  1  marble  toilet  bureau,  $42.00  ;  1  extension 
dining  table,  $45.00  ;  12  oak  dining  chairs  at  $3.50  ;  1  time- 
piece, $8.00 ;  and  1  whatnot,  $25.00. 

3.  John  Smith  sold  to  David  Brown  the  following  articles, 
viz.:  Oct.  7,  1854,  13  bushels  of  potatoes  at  $.56,  19  bu. 
com  at  $.97,  and  3  bbl.  of  apples  at  $2.42;  Nov.  1,  4  tons 
of  hay  at  $19,  and  3  tons  at  $18.50 ;  Dec.  17,  40  bushels 
of  potatoes  at  $.67,  34  bu.  of  corn  at  $.98,  and  21  bbl.  of 
apples  at  $2.75  per  barrel.  Jan.  1,  1855,  the  account  was 
settled  by  a  due  bill. 

Note.  —  A  due  bill  is  not  a  promise  to  pay  a  debt,  but  merely  an 
acknowledgment  that  it  is  due.  It  is  intended  to  cut  off  after  disputes 
as  to  the  debt  for  which  it  is  given,  by  furnishing  the  creditor  additional 
means  of  establishing  the  justice  of  his  claim.  Every  due  bill,  or  writ- 
ten promise  to  pay  money,  should  contain  the  words  "  value  received," 
to  show  that  the  person  who  signs  it  has  received  an  equivalent  for  it. 
Indeed,  it  is  a  principle  in  law,  that  no  claim  is  valid  unless  it  is  based 
on  some  service  rendered,  or  consideration  given,  to  the  person  against 
whom  it  is  made,  or  on  his  account. 

To  illustrate  the  form  of  a  due  bill,  we  will  suppose  that  John  Smith 
owes  James  Brown  $25,  and  that  he  gives  him  a  due  bill  for  the  amount, 
as  follows :  — 

/i'J.  ^o^f(m,   Jan.  /,  ^S55. 

cCoiia^^j    p^z    vatac    lecetved. 


132  BILLS    OF    GOODS. 

The  receipt,  when  a  due  bill  is  given,  might  be,  "  Received  payment 
by  due  bill,"  or,  "  Settled  by  due  bill." 

4.  May  7,  1855,  Hill  &  Saunders,  of  Boston,  sold  to  James 
Drew,  1  cask  linseed  oil,  24  gal.,  at  $1.50  per  gal. ;  1  bag 
Java  coffee,  122  lbs.,  at  16  cents  per  lb. ;  1  hhd.  of  N.  O. 
molasses,  127  gals.,  at  34  cents  per  gal. ;  1  chest  tea,  84  lb., 
at  48  cents  per  lb. ;  1  bag  pepper,  24  lbs.,  at  11  cents  per  lb. ; 

I  box  N.  O.  sugar,  278  lbs.,  at  6  cents  per  lb. 

5.  April  21,  1855,  French  &  Simmons,  of  New  York, 
sold  to  Clark  &  Hubbard,  1  piece  super,  broadcloth,  26  yds., 
at  $3.75  per  yd. ;  1  piece  fancy  cassimere,  31  yds.,  at  $1.34 
per  yd. ;  1  piece  black  cassimere,  23  yds.,  at  $1.62  per  yd. ; 
6  pieces  English  prints,  29,  31,  30,  33,  29,  and  31yds.,  or 
183  yds.,  at  $.12  per  yd. ;  3  pieces  Merrimack  prints,  28,  32, 
31  yds.,  or  91  yds.,  at  $.09  per  yd. ;  2  pieces  Scotch  gingham, 
42  and  41  yds.,  or  83  yds.,  at  $.17  per  yd. ;  1  piece  sarsenet 
cambric,  32  yds.,  at  $.28  per  yd. ;  3  dozen  cotton  hose  at  $1.87 
per  doz. ;  and  2  lbs.  Marshall's  linen  thread  at  $1.75  per  lb. 

6.  Geo.  Stevens,  of  Worcester,  sold  to  Daniel  Barnard, 
Sept.  15,  1854,  28  bu.  of  potatoes  at  $.87  ;  Oct.  1,  43  bu. 
of  potatoes  at  $.65  ;  Oct.  7,  7  tons  of  hay  at  $19.50,  4  tons 
at  $18.37,  and  2  tons  at  $17.25  ;  and  Nov.  3,  23  bbl.  of 
apples  at  $1.75,  and  19  bbl.  at  $1.94.  Mr.  Barnard  paid  Mr. 
Stevens,  Oct.  5,  1854,  $20,  Oct.  17,  $13.50;  and  Nov.  20, 
he  sold  him  14  lb.  sugar  at  7  cents,  12  lb.  at  9  cents,  8  lb.  at 

II  cents,  7  lb.  coffee  at  15  cents,  4  lb.  tea  at  54  cents,  3  lb.  at 
47  cents,  4  lb.  chocolate  at  19  cents,  and  1  bag  salt  for  $1.25. 
Jan.  1,  1855,  Mr.  Stevens  made  out  his  bill. 


PROPERTIES  OF  NUMBERS,  &C.  135 

SECTION  IX. 

PROPERTIES  OF  NUMBERS,  TESTS  OF  DIVISI- 
BILITY, FACTORS,  MULTIPLES,  DIVISORS. 

101.    Definitions. 

(«.)  A  FACTOR  of  any  given  number  is  such  a  number  as 
taken  an  entire  number  of  times  will  produce  the  given  num- 
ber ;  or,  the  factors  of  a  number  are  the  numbers  which 
multiplied  together  will  produce  it. 

Thus,  2  is  a  factor  of  4,  6,  8,  &c. ;  3  is  a  factor  of  6,  9,  12,  15,  &c. 

(5.)  A  DIVISOR  of  a  number  is  any  number  which  will 
exactly  divide  it. 

Note.  —  Every  divisor  of  a  number  must  be  a  factor  of  it,  and  every 
factor  of  a  number  a  divisor  of  it.  The  terms  factor  and  divisor,  as 
here  used,  are  cr.ly  applied  to  entire  numbers. 

(c.)  A  PRIME  NUMBER  is  ouc  which  has  no  other  factors 
besides  itself  and  unity. 

Thus.  1,  2,  3,  5,  7,  11,  13,  17,  &c.,  are  prime  numbers. 

(d.)  A  COMPOSITE  NUMBER  is  onc  which  has  other  fac- 
tors besides  itself  and  unity. 

Thus,  4,  6,  8,  9,  10,  12,  14,  15,  16,  &c.,  are  composite  numbers. 

(e.)  Any  entire  number  of  times  a  given  number  is  a 
MULTIPLE  of  it ;  or,  a  multiple  of  a  number  is  any  number 
which  can  be  exactly  divided  by  it. 

Thus,  12  is  a  multiple  of  1,  2,  3,  4,  6,  and  12,  because  it  is  an  exact 
number  of  times  each  of  them,  or  because  it  can  be  divided  by  each 
without  a  remainder. 

(y.)  Two  numbers  are  prime  to  each  other  when 
they  have  no  conmion  factor. 

For  example,  4  and  9  are  prime  to  each,  as  are  8  and  15,  24  and 
35,  &c. 

Again,  6  and  9  are  not  prime  to  each  other,  because  they  have  the 
common  factor  3 ;  8  and  12  are  not  prime  to  each  other,  because  they 
have  the  common  factor  4,  &c. 
12 


134  PBOPERTIES    OP  NUMBERS,    &C. 

Note.  —  It  is  obvious  from  the  foregoing,  that  every  number  is  a 
factor  of  all  its  multiples,  and  a  multiple  of  all  its  factors. 


109.    Demonstration  of  Principles, 

Proposition  First*  —  If  one  of  two  numbers  is  a  factor  of 
another,  it  must  he  a  factor  of  any  number  of  times  that  other 
number. 

For  to  find  any  number  of  times  a  given  number,  we  have 
only  to  multiply  the  number  by  some  new  factor,  without 
striking  out  any  of  the  former  ones. 

Illustrations.  —  Since  2  is  a  factor  of  12,  it  must  be  a  factor  of  any 
number  of  times  12,  as  24,  36,  48,  &c. 

Since  7  is  a  factor  of  14,  it  must  be  a  factor  of  any  number  of 
times  14,  as  28,  42,  56,  &c. 

Proposition  Second,  —  If  each  of  two  numbers  is  a  multi- 
ple of  a  third  number,  their  sum  and  their  difference  must 
also  be  midtiples  of  that  third  number. 

For,  adding  an  exact  number  of  times  a  given  number  to, 
or  subtracting  it  from,  an  exact  number  of  times  the  same 
number,  must  give  an  exact  number  of  times  that  number. 

Illustrations.  8  times  9,  or  72,  -}-  3  times  9,  or  27,  =  II  times  9,  or 
99.     So  8  times  9,  or  72,  —  3  times  9,  or  27,  =  5  times  9,  or  45. 

Again.  Both  12  and  20  are  multiples  of  4,  and  so  is  their  sum,  32, 
and  their  difference,  8. 

Both  42  and  28  are  multiples  of  7,  and  so  is  their  sum,  70,  and  their 
difference,  14. 

Proposition  Third.  —  If  one  of  two  numbers  is  a  multiple 
of  a  third  number,  and  the  other  is  not,  neither  their  sum  nor 
their  difference  will  be  a  multiple  of  that  third  number. 

For  both  the  sum  and  the  difference  of  an  entire,  and  a 
fractional,  number  of  times  a  given  number,  must  equal  a 
fractional  number  of  times  that  given  number. 

Illustrations.  8  times  6,  or  48,  +  2^  times  6,  or  15,  =  lOJ  times  6, 
or  63. 

8  times  6,  or  48,  —  2i.  times  6,  or  15,  =  5^  times  6,  or  33. 

7^  times  9,  or  66,  —  4  times  9,  or  36,  =  3|  times  9,  or  30. 

7^  times  9,  or  66,  +  4  times  9,  or  36,  =  11^  times  9,  or  102. 


PROPERTIES    OF   NUMBERS,   &C.  135 

Again.  20  is  a  multiple  of  5,  and  13  is  not;  hence,  neither  their 
sum,  33,  nor  their  difference,  7,  is  a  multiple  of  it. 

38  is  not  a  multiple  of  6,  and  24  is ;  hence,  neither  their  sum,  62, 
nor  their  difference,  14,  is  a  multiple  of  it. 

Proposition  Fourth.  —  If  neither  of  two  numbers  is  a  mul- 
tiple of  a  third,  their  sum  or  their  difference  may  or  may  not 
be  a  multiple  of  it. 

The  truth  of  this  proposition  can  best  be  made  manifest  by 
a  few  illustrations. 

1.  Neither  7  nor  23  is  a  multiple  of  2  ;  yet  both  their  sum,  30,  and 
their  difference,  16,  are  multiples  of  2. 

2.  Neither  5  nor  28  is  a  multiple  of  3 ;  yet  their  sum,  33,  is,  and 
their  difference,  23,  is  not,  a  multiple  of  3. 

3.  Neither  8  nor  17  is  a  multiple  of  3  ;  yet  their  sum,  25,  is  not,  and 
their  difference,  9,  is,  a  multiple  of  3. 

4.  Neither  14  nor  27  is  a  multiple  of  4 ;  and  neither  their  sum,  41, 
nor  their  difference,  13,  is  a  multiple  of  4. 

103.    Tests  of  the  Divisibility  of  Numbers. 

Application  of  the  foregoing  propositions. 

I.  Divisibility  by  2,  5,  3^,  or  by  any  other  number  which 
will  exactly  divide  10. 

Every  number  greater  than  ten  is  composed  of  a  certain 
number  of  tens,  plus  the  units  expressed  by  its  right  hand 
figure.  But  the  part  which  is  made  up  of  tens  must  (lOS, 
Prop.  I.)  be  divisible  by  any  divisor  of  ten  ;  and  hence, 
(lO^,  Prop.  II.  and  III.)  the  divisibility  of  the  entire  num- 
ber will  depend  on  the  part  expressed  by  the  right  hand 
figure. 

Therefore,  a  number  is  divisible  by  2,  5,  2  j,  1§,  or  by  any 
other  number  which  will  exactly  divide  10,  when  its  right  hand 
figure  is  thus  divisible. 

Ulttstration.  4125  is  divisible  by  5,  by  2i,  and  by  12,  because  each 
of  these  numbers  will  exactly  divide  10,  and  also  5,  the  right  hand  figure 
of  the  given  number. 

II.  Divisibility  by  4,  20,  25,  50,  12^,  16f,  or  any  other 
number  which  will  exactly  divide  100. 


136  PROPERTIES    OF    NUMBERS,   &C. 

Every  number  greater  than  100  is  composed  of  a  certain 
number  of  hundreds,  plus  the  number  expressed  by  its  two 
right  hand  figures.  But  the  part  which  is  made  up  of  hun- 
dreds must  (103,  Prop.  I.)  be  divisible  by  any  divisor  of 
one  hundred,  and  hence,  (lOS,  Prop.  II.  and  III.)  the  divisi- 
bility of  the  entire  number  must  depend  on  the  part  expressed 
by  the  two  right  hand  figures. 

Therefore,  a  number  is  divisible  by  4,  20,  25,  50,  12^,  or 
by  any  other  number  which  will  exactly  divide  100,  where 
its  two  right  hand  figures  are  thus  divisible. 

III.  Divisibility  by  8,  40,  125,  250,  500,  333^,  166|,  or  hj 
any  other  number  which  will  exactly  divide  1000. 

Every  number  greater  than  1000  is  composed  of  a  certain 
number  of  thousands,  plus  the  number  expressed  by  its  three 
right  hand  figures.  But  the  part  which  is  composed  of  thou- 
sands must  (103,  Prop.  I.)  be  divisible  by  any  divisor  of 
1000,  and  hence,  (lOS,  Prop.  II.  and  III.)  the  divisibility 
of  the  entire  number  must  depend  on  the  three  right  hand 
figures. 

Therefore  a  number  is  divisible  by  8,  125,  250,  500,  166|, 
333^,  or  by  any  number  which  will  exactly  divide  1000,  when 
its  three  right  hand  figures  are  thus  divisible. 

Note.  —  Similar  tests  for  determining  the  divisibility  of  numbers  by 
any  divisor  of  10,000,  100,000,  &c.,  could  be  established,  but  as  they 
could  very  rarely  be  applied  to  advantage,  we  omit  them. 

IV.  Divisibility  by  9. 

{a.)  If  1  be  subtracted  from  a  unit  of  any  decimal  denom- 
ination above  unity,  the  remainder  will  be  expressed  entirely 
by  9's,  and  will  therefore  be  a  multiple  of  9. 

Illustrations.  — 

10— 1  =9 

100  —  1  =  99 

1000  —  1  =  999 

10000  —  1  =  9999 

&c.,  &c. 

(5.)  But  unity,  orl,r=:0X9-|-l;  Jvnd  if,  for  convenience 
of  statement,  we  regard  0  X   9,  or  0,  as  a  multiple  of  9,  it 


PROPERTIES  OF  NUMBERS,  &;C.  137 

will  follow  that  a  unit  of  any  decimal  denomination  is  1  more 
than  a  multiple  of  9. 

Illustrations.  — 

1=0  +  1 

10  =  9  +  1 

100  =  99  +  1 

1000  =  999  +  1 

&c.,  &c. 

(c.)  As  a  unit  of  any  decimal  denomination  is  1  more  than 
u  multiple  of  9,  2  units  must  be  2  more  thair  SQch  a  multiple, 
3  units  must  be  3  more,  4  units  4  more,  5  units  5  more,  &c 

Illustrations.  —  Since  1000  =  1  more  than  a  multiple  of  9,  7000  must 
equal  7  more. 

Since  1000000  =  1  more  than  a  multiple  of  9,  7000000  must  equal 
7  more,  &c. 

(d.)  But  the  digit  figures  of  a  number  express  the  number 
of  units  of  its  various  denominations,  and  therefore  any  num- 
ber must  be  as  many  more  than  a  multiple  of  9  as  there  are 
units  in  the  sum  of  its  digit  figures. 

Illustrations.      8235  =  8000  +  200  +  30  +  5. 

8000  =  8  more  than  a  multiple  of  9. 

200  =  2  more  than  a  multiple  of  9. 

30  =  3  more  than  a  multiple  of  9. 

5  =  5  more  than  a  multiple  of  9. 

Therefore,  8235  =  8  +  2  +  3  +  5,  or  18  more  than  a  multiple  of  9, 
or  it  equals  a  multiple  of  9,  plus  18,  and  must  therefore  (since  9  is  a 
divisor  of  18)  be  a  multiple 'of  9. 

Again.     57864  =  50000  +  7000  +  800  +  60  +  4. 

50000  =:  5  more  than  a  multiple  of  9. 

7000  =  7     "         "     "         "        "    " 

800  =  8     "         "     "         "         "    " 

60  =  6     ^'         "      ''         "         "    " 

4  =  4"         «      "         "         "    " 


Therefore,  57864  =  5  +  7  +  8  +  6  +  4,  or  30  more  than  a  multi- 
ple of  9,  and  is  therefore  (since  9  is  not  a  divisor  of  30)  not  a  multiple 
of  9. 

(c.)   From  these  principles  it  follows,  — 
1.    Tliat  every  number  is  equal  to  a  multiple  of  9,  plus  the 
turn  of  its  digit  figures. 
12* 


138  PROPERTIES    OP   NUMBERS,    &C. 

2.  That  if  the  sum  of  the  digit  figures  of  any  number  he 
subtracted  from  it,  the  remainder  will  be  a  multiple  of  9. 

3.  That  a  number  is  divisible  by  9  when  the  sum  of  its 
digit  figures  is  thus  divisible. 

4.  That  the  remainder  obtained  by  dividing  the  sum  of  the 
digit  figures  of  any  number  by  9,  is  the  same  as  that  obtained 
by  dividing  the  number  itself  by  9. 

5.  TJiat  the  difference  of  any  two  numbers,  the  sums  of 
whose  digits  are  alike,  will  be  a  multiple  of  9. 

6.  That  the  divisibility  of  a  number  by  9  will  not  he  affect^ 
ed  by  any  change  in  the  order  of  its  digits. 

7.  That  if  the  digit  figures  of  a  number  he  added  together, 
and  then  the  digit  figures  of  the  result,  and  so  on  till  the  sum 
is  expressed  by  a  single  figure,  that  figure  will  either  be  9,  or 
the  remainder  obtained  by  dividing  the  original  number  by  9. 
If  the  figure  is  9,  the  number  is  a  multiple  of  9. 

Remark.  —  rinding  the  excess  of  any  number  over  a  multiple  of  9 
is  called  casting  out  the  9's. 

(f.)  Consequences  of  the  foregoing.  —  From  the  foregoing  properties 
of  the  number  9,  considered  in  connection  with  the  principles  established 
in  1 02,  come  some  convenient  methods  of  proving  numerical  opera- 
tions ;  a  few  of  which  we  will  mention,  leaving  the  pupil  to  find  out  the 
reasons  for  each. 

1.  To  prove  Addition.  —  Cast  out  the  9's  from  the  several  numbers 
added,  add  the  results,  and  cast  out  the  9's  from  their  sum.  Then  cast 
out  the  9's  from  the  number  obtained  as  the  answer  to  the  question,  and 
if  the  work  be  correct,  the  last  two  results  will  be  equal. 

2.  To  prove  Subtraction.  —  Cast  out  the  9's  from  the  subtrahend  and 
remainder,  add  the  results,  and  cast  out  the  9's  from  their  sum.  Then 
cast  out  the  9's  from  the  minuend,  and  if  the  work  is  coiTect,  the  last 
two  results  will  be  equal. 

3.  To  prove  MultipUcation.  —  Cast  out  the  9's  from  the  several  factors 
employed,  multiply  the  results  together,  and  cast  out  the  9'8  from 
their  product.  Then  cast  out  the  9'8  from  the  product  of  the  original 
multiplication,  and  if  the  work  is  correct,  the  last  two  results  will  be 
equal. 

4.  To  prove  Division.  —  Cast  out  the  9's  from  the  divisor,  quotient, 
and  remainder ;  to  the  product  of  the  first  two  results  add  the  last 
result,  and  cast  out  the  9's.  Then  cast  out  the  9's  from  the  dividend, 
and  if  the  work  is  correct,  the  last  two  results  will  be  equal. 


PROPERTIES    OF    NUMBERS,   &C.  139 

(g.)  Dependent  on  the  same  principles  is  the  following,  which  tho 
pupil  raay  use  ivith  the  uninitiated  as  an  arithmetical  puzzle. 

To  tell  what  figure  has  been  erased.  —  Tell  a  person  to  write  any  num- 
ber whatever,  without  informing  you  what  it  is  ;  to  subtract  the  sum  of 
its  digit  figures  from  it ;  to  erase  from  this  result  any  digit  figure,  other 
than  zero,  and  write  zero  in  its  place ;  and  finally,  to  add  together  the 
digit  figures  of  the  number  thus  obtained,  and  tell  you  their  sum. 

The  difference  between  this  sum  and  the  next  higher  multiple  of  9 
will  show  the  figure  removed.  Thus,  if  the  sum  is  29,  7  was  erased  ;  if 
it  is  45,  9  was  erased  ;  &c. 

Let  the  pupil  explain  the  reasons  of  this. 

V.    Divisihility  hy  3. 

Since  9  is  a  multiple  of  3,  every  number  which  is  a  multi- 
ple of  9  must  also  be  a  multiple  of  3,  (Prop.  I.)  Therefore, 
a  unit  of  any  decimal  denomination  must  be  1  more  than  a 
multiple  of  3  ;  and  hence  a  number  is  a  multiple  of  3  when 
the  sum  of  its  digit  figures  is  such  a  multiple. 

YI.    Divisibility  by  11, 

(a.)  A  unit  of  any  decimal  denomination  is  either  1  or  10 
more  than  a  multiple  of  11.     Thus, — 

l=OX  11  +1 

10  =  0  X  11  -i-  10 

100  =  9  X  11  -f  1 

1000  =  90  X  11  +  10 

10000  =  909  X  11  +  I 

100000  =  9090  X  11  4-  10 

Or,  arranging  the  numbers  with  reference  to  the  remainders,  we 
have  — 

1=0  X  11  +  1 
100  =  9  X  11  4-1 
10000  =  909  X  11  +1  - 
1000000  =  90909  X  1 1  -f  1 

10  =  0X11  +  10  =  1X11  —  1 

1000  =  90  X  11  +  10  =  91  X  11  —  1 
100000  =  9090  X  11  +  10  =  9091  X  H  —  1 
10000000  =  909090  X  1 1  -(-  10  =  909091  X  1 1  —  1 , 

[b.)  From  which  we  see  that  a  unit  of  any  denomination 
expressed  by  a  figure  occupying  the  1st,  3d,  5th,  or  any  other 
odd  place  from  the  point,  is  1  more  than  a  multiple  of  11  ; 
and  that  a  unit  of  any  denomination  expressed  by  a  figure 


140  PROl'KRTIKS    OF    NUMBERS,   &C. 

occupying  the  2d,  4th,  or  any  other  even  place  from  the 
point,  is  1  less  than  a  multiple  of  11. 

(c.)  Hence,  on  account  of  the  figures  occupying  odd  places 
from  the  point,  a  number  is  as  many  more  than  a  multiple  of 

11  as  there  are  units  in  the  sum  of  these  figures,  while  on 
account  of  the  figures  occupying  even  places,  it  is  as  many 

,less  than  a  multiple  of  11  as  there  are  units  in  the  sum  of 
those  figures. 

id.)  Hence,  every  number  is  equal  to  some  multiple  of  11, 
plus  the  sum  of  its  digit  figures  occupying  odd  places  from 
the  point,  minus  the  sum  of  those  occupying  even  places.  If 
these  sums  are  alike,  the  additions  will  equal  the  subtractions, 
and  the  number  will  be  a  multiple  of  11.  If  these  sums  are 
unlike,  their  difference  will  be  the  excess  of  the  additions 
over  the  subtractions,  or  of  the  subtractions  over  the  ad- 
ditions. 

(e.)  If,  then,  the  difference  of  the  sums  of  the  alternate 
digits  is  a  multiple  of  11,  the  whole  number  will  be  either 
the  sum  or  the  difference  of  two  multiples  of  11,  and  hence  a 
multiple  of  11  ;  but  if  this  difference  is  not  such  a  multiple, 
the  whole  number  will  be  either  the  sum  or  the  difference  of 
two  numbers,  one  of  which  is,  and  the  other  is  not,  a  multiple 
of  11,  and  hence  will  not  be  a  multiple  of  11. 

(yi)  Hence,  a  number  is  a  multiple  o/"  11,  when  the  sums 
of  its  alternate  digits  are  equal,  or  when  their  difference  is  a 
multiple  of  11. 

First  Example.  —  Is  15873  a  multiple  of  11  ? 

Solution. —  The  sum  of  the  digits  in  tlie  odd  places  is  3  +  8  -f-  1.  or 

12  ;  the  sum  of  those  in  the  even  places  is  7  -f-  5i  or  12".  Hence,  the 
two  sums  are  alike,  and  15873  is  a  multiple  of  11.     v^ 

Second  Example.  —  Is  274854  a  multiple  of  1 1  ? 

Solution.  —  The  sum  of  the  digits  in  the  odd  places  is  4  -f  8  -f  7  = 
19 ;  the  sura  of  the  digits  in  the  even  places  is5-f-4-f-2  =  ll;  the 
difference  between  19  and  11  is  8,  which  is  not  a  multiple  of  11.  Henc«^ 
274854  is  not  a  multiple  of  11. 

VII.    If  a  number  is  divisible  by  each  of  two  numbers 


PROPERTIES    OF   NUMBERS,  &C.  141 

which  are  prime  to  each  other,  it  will  be  divisible  by  their 
product. 

For  dividing  by  one  cannot  (since  the  numbers  are  prime 
to  each  other)  cast  out  the  other,  or  any  factor  of  it. 

Illustrations.  1.  A  number  which  is  divisible  by  4  and  9  must  b3 
divisible  by  4  X  9,  or  36. 

2.  A  number  which  is  divisible  by  8  and  15  must  be  divisible  by  8  X 
15,  or  120 

3.  A  number  is  divisible  by  12,  when  it  is  by  3  and  4. 

4.  A  number  is  divisible  by  35,  when  it  is  by  7  and  5. 

5.  A  number  is  divisible  by  42,  when  it  is  by  6  and  7. 

Vni.  If  a  number  is  divisible  by  each  of  two  numbers 
which  have  a  common  factor,  it  will  not  of  necessity  be  divisi' 
hie  by  their  product. 

For  dividing  by  one  must  cast  out  the  common  factor  of 
the  two  numbers,  and  if  that  factor  be  not  taken  more  than 
once  as  a  factor  of  the  original  number,  the  quotient  will  not 
be  divisible  by  the  other  of  the  two  numbers. 

Illustrations.  84  is  divisible  by  4  and  6,  but  not  by  their  product,  24. 
72  IS  divisible  by  4  and  6,  and  also  by  their  product,  24. 

IX.  If  one  number  is  not  divisible  by  another,  it  will  not 
he  divisible  by  any  multiple  of  that  other  number. 

For  if  a  number  ^does  not  contain  once  another  number,  it 
cannot  contain  any  number  of  times  that  other  number. 

Illustrations.  —  A  number  which  is  not  divisible  by  2.  is  not  divisible 
by  4,  6,  8,  10,  &c.  A  number  which  is  not  divisible  by  3,  is  not  divisi- 
ble by  6,  9,  12,  15,  18,  21,  &c. 

X.  A  number  which  ia  divisible  by  any  composite  number 
is  divisible  by  all  the  factors  of  that  composite  number. 

For  dividing  by  any  composite  number  is  merely  dividing 
by  the  product  of  its  factors. 

104.    Recapitulation,  for  convenience  of  reference. 

I.  Any  number  is  divisible  by  2,  5,  3^,  or  any  other  number  which  will 
exactly  divide  10,  when  its  right  hand  figure  is  thus  divisible. 

II.  Any  number  is  divisible  by  4,  20,  25,  50,  12i.,  162,  or  any  other 
number  which  will  exactly  divide  100,  when  the  number  expressed  by  its  two 
right  hand  figures  is  thus  divisible. 


142  PROPERTIES  OF  NUMBERS,  &C. 

m.  An^  number  is  divisible  by  8,  4,  125,  250,  3331,  or  any  other  nwrn- 
her  which  will  exactly  divide  1000,  when  the  number  expressed  by  its  three 
right  hand  Jigures  is  thus  divisible. 

TV.  Any  number  is  divisible  by  3  or  9,  when  the  sum  of  its  digits  is  thm 
divisible. 

V.  Any  number  is  divisible  6y  11,  when  the  sums  of  its  alternate  digits 
are  equal,  or  their  difference  is  a  multiple  of  11. 

VI.  A  number  which  is  divisible  by  each  of  two  numbers  which  are  primf 
to  each  other,  is  divisible  by  their  product. 

VII.  A  number  which  is  divisible  by  each  of  two  numbers  not  prime  tc 
each  other,  is  not  of  necessity  divisible  by  their  product. 

VIII.  A  number  which  is  not  divisible  by  another  is  not  divisible  by  any 
multiple  of  that  other  number. 

IX.  A  number  which  is  divisible  by  any  composite  number  is  also  divisi- 
ble by  all  the  factors  of  that  composite  number. 

10^.    Definittons  of  Factors^  Powers,  S^c. 

(a.)  A  number  is  said  to  be  divided  into  factors  when  any 
factors  which  will  produce  it  are  found. 

Thus,  in  36  =  4  X  9,  36  is  divided  into  the  factors  4  and  9 ;  but  in 
S6  =  2  X  6  X  3  it  is  divided  into  the  factors  2,  6,  and  3. 

(b.)    A  number  is  said  to  be  divided  into  its  prime  factors 
when  it  is  divided  into  factors  which  are  all  prime  numbers. 
Illustrations.       36  =  2X2X3X3  8  =  2X2X2 

30  =  2X3X5  84  =  2X2X3X7 

(c.)  When  any  number  is  taken  more  than  once  as  a  factor 
to  produce  another  number,  we  may  express  the  number  of 
times  it  is  taken  as  a  factor,  by  placing  a  small  figure  above 
it  and  a  little  to  the  right. 

Illustrations.  3^  means  the  same  as  3  X  3  ;  i.  e.,  that  3  is  to  be  taken 
twice  as  a  factor. 

3*  means  the  same  as  3  X  3  X  3  X  3 ;  i.  e.,  that  3  is  to  bo  taken  4 
times  as  a  factor. 

3»  X  2*  means  the  same  as  3X3X3X2X2X2X2;  i.  e.,  that 
the  product  of  3  taken  3  times  as  a  factor  is  to  be  multiplied  by  the 
product  of  2  taken  4  times  as  a  factor. 

Note.  —  The  student  should  notice  the  difference  between  taking  a 
number  as  a  factor  a  certain  number  of  times,  and  merely  taking  it  a 
number  of  times. 


METHOD    OF    FACTORING    NUMBERS.  143 

Thus,  5  taken  3  times  as  a  factor  =  5  times  5  times  5  =  125,  but  5 
takm  3  times  =  3  times  5  =  15. 

{d.)  The  product  of  a  number  taken  any  number  of  times 
as  a  factor  is  called  a  power  of  the  number,        ^ 

Illustrations.  9  is  the  second  power  of  3,  because  it  is  the  product  of 
32,  i.  e.,  of  3  taken  twice  as  a  factor. 

25  is  the  fifth  power  of  2,  because  it  is  the  product  of  2*  i.  e.,  of  2 
taken  5  times  as  a  factor. 

(e.)  The  figure  indicating  how  many  times  a  number  is 
taken  as  a  factor  is  called  the  exponent  of  the  power  to 
which  the  number  is  raised. 

Thus,  in  2^,  the  exponent  is  5  ;  in  5^,  it  is  2  ;  in  6^,  it  is  4  ;  &c. 

(f)  The  following  examples  indicate  the  method  of  read- 
ing numbers  expressing  powers. 

3^  is  read  three  seventh  power.,  or  three  to  the  seventh  power. 
45  X  2^  is  read  four  fifth  power  multiplied  by  two  third  power. 

Note.  —  The  second  power  of  a  number  is  sometimes  called  itr 
SQUAKE,  and  the  third  power  its  cube.  / 

100.    Method  of  Factoring  Numbers. 

1.  What  are  the  prime  factors  of  2772  ? 

Solution.  —  We  see  by  104,  II.,  that  2772  is  divisible  by  4,  and 
therefore  by  2  X  2,  the  prime  factors  of  4. 

Dividing  by  4  gives  693  for  a  quotient,  which,  by  104,  IV.,  we  see 
is  divisible  by  9,  and  therefore  by  3X3,  the  prime  factors  of  9. 

Dividing  693  by  9  gives  77  for  a  quotient,  the  prime  factors  of  which 
are  7  and  11.  Hence,  the  prime  factors  of  2772  are  2^,  32,  7,  and  11 ; 
or,  which  is  the  same  thing,  2772  =  22  X  32  X  7  X  11. 

2.  Wiiat  are  the  prime  factors  of  29766  ? 

Solution.  —  We  see  by  104,  I.  and  IV.,  that  29766  is  divisible  by 
both  2  and  3,  and  hence  by  their  product,  6.  Dividing  by  6  gives  4961 
for  a  quotient,  which,  by  104,  V.,  is  a  multiple  of  11.  Dividing  by  U 
gives  451  for  a  quotient,  which  (104,  V.)  is  also  a  multiple  of  11 
Dividing  by  11  gives  41  for  a  quotient,  which  is  obviously  a  prime 
number.     Therefore,  29766  =  2  X  3  X  11^  X  41. 

3.  What  are  the  prime  factors  of  3871123  ? 

Solution.  —  We  readily  see  that  it  is  not  divisible  by  2,  3,  5,  or  11, 


144  FAcrroRS  op  numbers. 

and  therefore  we  must  see  whether  it  is  divisible  by  the  other  prime 
numbers,  beginning,  for  convenience  sake,  with  the  smallest.  By  actaal 
trial,  we  find  that  7,  13,  17,  19,  and  23  each  leave  a  remainder  after 
division,  but  that  29  is  contained  in  it  exactly  133487  times.  Therefore, 
29  is  one  of  the  prime  factors  of  the  given  number,  and  the  others  will 
be  found  in  133487.  But  no  number  less  than  29  can  be  a  factor  of 
133487,  for  none  is  a  factor  of  the  original  number.  We  therefore  first 
try  29,  wliich  we  find  is  contained  exactly  4603  times.  Therefore,  29  is 
again  a  factor  of  the  original  number,  and  the  remaining  factors  must 
be  found  in  4603.  But  no  number  less  than  29  can  be  a  factor  of  4603, 
for  none  was  a  factor  of  the  original  number.  Beginning  with  29,  we 
try  in  succession  29,  31,  37,  41,  43,  47,  53,  59,  61,  67,  and  71,  and  find 
that  none  of  them  will  exactly  divide  4603  ;  and  we  observe,  moreover, 
that  the  entire  part  of  the  quotient  of  the  division  by  71  is  less  than  71. 

Now,  as  the  divisor  increases,  the  quotient  mu;^t  decrease ;  therefore, 
if  there  is  any  number  larger  than  71  which  will  divide  it,  the  quotient 
must  be  less  than  71.  But  when  a  division  can  exactly  be  performed, 
the  divisor  and  quotient  must  each  be  a  factor  of  the  dividend ;  and 
hence  if  4603  has  a  factor  greater  than  71,  it  must  also  have  one  less 
than  71.  But  as  we  have  already  found  that  it  has  no  factor  less  than 
71,  we  infer  that  it  can  have  none  greater,  aud  that  it  is  a  prime  number. 

Hence,  3871123  ==  292  x  4603. 

(a.)  The  labor  of  testing  the  divisibility  of  a  number  by 
the  various  prime  numbers  may  be  made  much  less  laborious 
ar.d  tedious,  by  considering  what  must  be  the  last  figure  of 
the  quotient,  if  the  division  by  any  number  is  possible. 

(6.)  Thus,  in  the  last  example,  in  testing  the  divisibility  of  4603,  we 
may  observe  that  if  29  is  a  factor  of  it,  the  last  figure  of  the  other  factor 
must  be  7,  for  7,  or  some  number  ending  in  7,  is  the  only  number  which, 
multiplied  by  29,  will  give  for  a  product  a  number  ending  in  3.  Divid- 
ing by  29,  we  have,  29  is  contained  in  46  once,  and  17  remainder;  in 
170,  5  times,  and  25  remainder;  and  we  can  readily  see  that  it  is  con 
tained  in  253  more  than  7  times. 

(c.)  Again.  If  31  is  a  factor  of  4603,  the  last  figure  of  the  other 
factor  must  be  3,  for  3,  or  some  number  ending  in  3,  is  the  only  number 
which,  multiplied  by  31,  will  give  for  a  product  a  number  ending  in  3. 
Dividing  by  31,  we  have,  31  is  contained  in  46  once,  and  15  remainder; 
in  150,  4  times,  and  26  remainder ;  and  we  can  see  at  a  glance  that  it  is 
contained  in  263  more  than  3  times. 

(rf.)  Again.  If  37  is  a  factor  of  4603,  the  last  figure  of  the  other 
factor  must  be  9.  But  37  is  contained  in  46  once,  and  9  remainder;  in 
90,  twice,  and  16  remainder;  and  it  is  obvious  that  it  cannot  be  con 
tained  as  ma&y  as  9  times  in  163. 


FACTORS    OF    NUx^ilBERS. 


145 


(e.)  From  the  above,  we  see  that  to  find  the  prime  factors 
of  a  number,  we  may  first  divide  it  by  any  number  which  will 
divide  it  without  a  remainder,  then  divide  this  quotient  by  any 
number  which  will  divide  it  without  a  remainder,  and  so  pro- 
C3ed  with  each  successive  quotient,  till  we  reach  one  which  is 
a  prime  number,  or  which  can  be  readily  divided  into  its 
prime  factors.  The  prime  factors  of  the  several  divisors  and 
of  the  last  quotient  will  be  the  prime  factors  required. 

(/.)  If  in  any  case  w^e  cannot  discover  a  divisor  of  any 
given  number  by  the  tests  of  104,  we  try  in  succession  7, 
and  all  the  prime  numbers  from  13  upwards,  till  we  find  one 
"which  is  a  divisor  of  the  given  number,  or  till  the  entire  part 
of  our  quotient  is  less  than  the  number  employed  as  a  divisor, 
in  which  case  the  number  is  prime. 


10T«    Exercises  for  the  Student. 

(a.)    Find  the  prime  factors  of  the  numbers  from  1  to  100, 
writing  them  as  in  the  following  model. 

1,  prime. 

2,  prime. 


3,  prime. 

4  =  2  X  2  =  2^8. 

5,  prime. 

6  =  2X3. 


.7,  pnme. 

8=2X2X2= k 

9  =  3  X  3  =  32. 
10  =  2  X  5. 
11,  prime. 
12  =  2  X  2  X  3  =  22  X  3. 


ih.)  We  would  recommend  that  the  pupil  make  out  a  table 
of  the  prime  factors  of  the  numbers  from  1  to  1000.  He 
will  find  it  a  very  profitable  exercise,  and  one  which  will 
greatly  aid  him  in  all  his  subsequent  work. 

{c.)    What  are  the  prime  factors  — 


1. 

Of  1001 ? 

2. 

Of  1025  ? 

3. 

Of  1024? 

4. 

Of  1033  ? 

5. 

Of  1096? 

€. 

Of  1157? 

7. 

Of  1067  ? 

8. 

Of  1183? 

9. 

Of  1625  ? 

10. 

Of  2057  ? 

11. 

Of  16128? 

12. 

Of  3809  ? 

13. 

Of  6381 ? 

14, 

Of  7128? 

13 


146 


DIVISORS  OF  NUMBERS. 

15. 

Of  7854  ? 

21. 

Of  444528  ? 
Of  1072181? 

16. 

Of  5989  ? 

22. 

17. 

Of  5625  ? 

23. 

Of  5764801 ? 

18. 

Of  9257 .? 

24. 

Of  6103515625? 

19. 

Of  10917  ? 

25. 

Of  12^68^87,046 

20. 

Of  843479  ? 

108.     Common  Divisor,  Definitions ,  and  Properties. 

(«.)  A  DIVISOR  of  a  number  is  a  number  which  will  ex- 
actly divide  it. 

(h.)    A   COMMON    DIVISOR    OF    TWO    OR    3I0RE    NUMBERS 

is  a  number  which  is  a  divisor  of  each  of  them. 

(c.)     The      GREATEST     COMMON     DIVISOR     OF     TWO      OR 

MORE  NUMBERS  is  the  largest  number  which  is  a  divisor  of 
each  of  them. 

(c?.)  From  these  definitions  and  the  jmnciples  previously 
established,  it  follows  that,  — 

1.  A  divisor  of  a  number  can  contain  only  stich  prime  fac» 
tors  as  are  found  in  that  number. 

2.  A  common  divisor  of  two  or  more  numbers  can  contain 
only  such  prime  factors  as  are  common  to  all  the  numbers. 

3.  The  greatest  common  divisor  of  two  or  more  numbers  is 
the  product  of  all  the  prime  factors  common  to  all  the  given 
numbers. 

109.     Greatest  Common  Divisor.  —  Method  by  Factors, 

1.  What  is  the  greatest  common  divisor  of  819  and  1071  ? 
Solution. —  The   greatest  common  divisor   of  819  and   1071   is  the 

product  of  all  the  prime  factors  common  to  those  numbers. 
819  =  32  X  7  X  13 
1071  =32  X  7  X  17 
From  which  we  see  that  the  only  common  factors  are  3''  and  7.     There- 
fore, 32  X  7,  or  63,  is  the  greatest  common  divisor  required. 

What  is  the  greatest  common  divisor  — 

2.  Of  792  and  936  ?     1   3-  Of  1125  and  1575  ? 


DIVISORS    OF    NUMBERS.  147 


4.  Of  3102  and  3666  ? 

5.  Of  287  and  369  ? 


6.  Of  4652  and  5544? 

7.  Of  924  and  1188? 


\ 


8.  What  is  the  greatest  common  divisor  of  72,  108,  and 
252? 

Solution.  —  The  greatest  common  divisor  of  72,  108,  and  252,  is  the 
product  of  all  the  prime  factors  common  to  those  numbers. 
72  =  23  X  32 
108  =  22  X  33 
252  =  22  X  32  X  7 
From  which  we  see  that  the  only  common  factors  are  22  and  32.    There- 
fore, 32  X  22,  or  36,  is  the  greatest  common  divisor  required. 

What  is  the  greatest  common  divisor  — 

9.  Of  168,  505,  and  539? 

10.  Of  1386,  3234,  and  4158  ? 

11.  Of  864,  2058,  and  2346  ? 

12.  Of  686,  1029,  and  2401  ? 

13.  Of  112,  147,  168,  and  189  ? 

14.  Of  576,  672,  864,  and  1132  ? 

110.    A  more  brief  Method. 

(a.)  The  following  solution  will  usually  be  found  much 
more  brief  than  the  preceding,  inasmuch  as  it  avoids  the 
necessity  of  separating  all  the  numbers  into  their  factors.  By 
it,  we  find  the  factors  of  any  one  of  the  numbers,  and  then  see 
which  of  them  are  factors  of  all  the  other  numbers. 

1.  What  is  the  greatest  common  divisor  of  756,  840,  1386, 
and  1596? 

Solution.  —  We  first  find  the  factors  of  840,  because  they  can  be  the 
most  readily  found.  840  =  2^  X  3  X  5  X  7,  and  we  have  now  to  find 
which  of  these  factors,  if  any,  are  factors  of  all  the  other  given  num- 
bers. It  is  obvious  (104,  I.)  that  2  is  a  factor  of  all  of  them,  and 
(104,  II.)  that  it  is  contained  but  once  as  a  factor  in  1386.  Hence,  2 
will  enter  once  only  as  a  factor  of  the  greatest  common  divisor.  5  is 
(104,  I.)  a  factor  of  no  other  number.  3  being  a  factor  (104,  IV.) 
of  all  the  numbers,  is  a  factor  of  the  greatest  common  divisor.  By 
trial,  7  is  found  to  be  a  factor  of  all  the  numbers,  and  hence  of  the  grear- 
ea*  common  divisor.  Hence,  the  greatest  common  divisor  is  2  X  3  X 
7.  or  42. 


148  DIVISORS    OF    Nl.M;;KiCS. 

(b.)    What  is  the  greatest  'X):nuion  Jiv'sor  — 

2.  Of  3465,  4375,  and  5  >ir)0  r' 

3.  Of  1792,  936,  1224,  md  1G56? 

4.  Of  1342,  1738,  2371/  and  2590  ? 

5.  Of  3312,  6048,  957G,  and  633G  ? 

6.  Of  4572,  2380,  5272,  and  83G4  ? 

7.  Of  3125,  4379,  8243,  and  5975  ? 

111.    Factoring  not  always  necessary. 

(a.)  "We  can  frequently  find  the  greatest  common  divisor 
of  the  given  numbers  without  finding  their  prime  factors. 

(6.)  Thus,  in  finding  the  greatest  common  divisor  of  24  and  42,  wo 
can  see  at  a  glance  that  6  is  a  divisor  of  both ;  that  24  =  6  X  4,  and 
42  =  6  X  7  ;  therefore,  since  4  and  7  have  no  common  factor,  6  must 
be  the  greatest  common  divisor  required. 

(c.)  Again.  In  finding  the  greatest  common  divisor  of  693  and  819 
we  see  at  once  that  9  will  divide  each. 

693  =  9  X  77 
819  =  9  X  91 
Now,  comparing  77  and  91,  we  see  that  7  will  divide  each. 

77  =  7  X  11 
91  =7  X  13 

And  as  7  and  11  are  prime  to  each  other,  there  is  no  other  common  fac- 
tor to  the  given  numbers.  Hence,  7  X  9,  or  63,  is  the  greatest  common 
divisor  required. 

{d.)  Again.  In  finding  the  greatest  common  divisor  of  36  and  72, 
we  may  see  that  36  is  a  divisor  of  72,  and  hence  the  greatest  ctdidmon 
divisor  required. 

(e.)    What  is  the  greatest  common  divisor  — 

1.  Of  12  and  18? 

2.  Of  28  and  42  ? 

3.  Of  18  and  54  ? 

4.  Of  48  and  84.? 

5.  Of  324  and  594? 

6.  Of  2025  and  2916? 

7.  Of  5544  and  11583? 

8.  Of  18,  48,  72,  and  66  ? 

9.  Of  12,36,  60,  and  132? 
10.  Of  49,  63,  84,  and  91  ? 


f 


.    DIVISORS    OF    NUMBERS.  149 

11^.    Method  by  Addition  and  Subtraction, 

(a.)  When  the  sum  or  the  difference  of  any  two  of  the 
given  numbers  is  a  small  number,  or  one  more  easily  divided 
into  factors  than  any  of  the  original  numbers,  the  work  may 
be  abbreviated  by  applying  the  principles  of  103?  Propo- 
sition 11. 

1.  What  is  the  G.  C.  D.*  of  8379  and  8484  ? 
Suggestion.  —  The  G.  C.  D.  required  must  be  a  divisor  of  8484  — 

8379,  which  is  105  =  3X5X7;  and  we  have  only  to  ascertain  which 
of  these  factors,  if  any,  are  factors  of  one  of  the  given  numbers. 

2.  What  is  the  G.  C.  D.  of  89437,  95429,  and  90537  ? 

Suggestion.  —  The  G.  C.  D.  required  must  be  a  divisor  of  the  differ- 
ence of  any  two  of  the  given  numbers,  as  of  89437  and  90537,  which  is 
1100  =  22  X  52  X  11 ;  and  we  have  only  to  ascertain  which  of  these 
factors,  if  any,  are  factors  of  all  the  given  numbers. 

3.  What  is  the  G.  C.  D.  of  56474  and  28526  ? 

Suggestion.  —  The  G.  C.  D.  required  must  be  a  divisor  of  56474  -f- 
28526,  which  is  85000  =  2^  X  5^  X  17 ;  and  we  have  only  to  ascertain 
which  of  these  factors,  if  any,  are  factors  of  one  of  the  given  numbers. 

4.  What  is  the  G.  C.  D.  of '3598,  5383,  6345,  and  8617  ? 
Suggestion.  —  The  G.  C.  T>.  required  must  be  a  divisor  of  the  sum  of 

any  two  of  the  given  numbers,  as  of  5383,  and  8617,  which  is  14000  := 
2*  X  53  X  7  ;  and  we  have  only  to  ascertain  which  of  these  factors,  if 
any,  are  factors  of  all  the  given  numbers. 

What  is  the  greatest  common  divisor  — 

5.  Of  949  and  962? 

6.  Of  857637  and  857692? 

7.  Of  5489  and  7689  ? 

8.  Of  9709,  10906,  and  10241  ? 

9.  Of  71227,  72553,  73840,  and  75127  ? 

10.  Of  930069,  992673,  and  1103673  ? 

11.  Of  12551  and  25949  ? 

12.  Of  169881  and  170119? 

13.  Of  16181  and  16324? 

14.  Of  180006,  293694,  468963,  and  596862  ? 

*  G.  C,  D.  means  greatest  common  divisor. 

13* 


150  DIVISOUS    OF    NUMBERS. 

113.    Demonstration  of  Method  hy  Division, 

(a.)  Proposition.  —  The  G.  G.  D.  of  any  two  numbers  is 
the  same  as  the  G.  G.  D.  of  the  smaller^  and  the  remainder 
left  after  dividing  the  larger  hy  the  smaller. 

{h.)  To  prove  this,  let  the  letter  a  represent  any  number  whatever^ 
and  the  letter  h  represent  any  other  number  larger  than  a.  Now,  what-- 
ever  numbers  a  and  b  represent,  it  is  obvious  that  a  is  contained  in  b 
some  number  of  times,  which  we  will  call  c  times,  and  that  there  may, 
or  may  not,  be  a  remainder.  If  there  is  no  remainder,  a  will  be  the 
G.  C.  D.  of  a  and  6.  If,  however,  there  is  a  remainder,  we  will  call  it  J, 
and  we  have  to  prove  that  the  G.  C.  D.  of  a  and  h  is  the  same  number 
as  the  G.  C.  D.  of  a  and  d. 

(c.)    Representing  the  division  by  writing  the  letters  as  we  should  the 
numbers  they  represent,  we  have,  — 
Dividend. 
Divisor  a  )    b    {  c  =  Quotient. 

a  X  c      =  Product  of  divisor  by  quotient. 

d  =  Remainder. 

{d.)   From  the  nature  of  division,  it  follows  that  — 
d  =  b  —  a  X  c 
and  b  =d -^  a  X  c 

(e.)  Now,  as  any  divisor  of  a  must  {1.02 j  Prop.  I.)  be  a  divisor  of 
c  X  «,  the  G.  C.  D.  sought  must  be  a  divisor  of  6  a;id  c  X  a,  and  hence 
(lOa,  Prop.  II.)  of  6  —  c  X  a,  or  d.  Again.  Any  divisor  of  d  and  a 
must  be  a  divisor  of  d  and  c  X  «,  and  hence  (102,  Prop.  II.)  of  d  -\- 
c  X  a,  or  b.  It  therefore  follows  that  the  G.  C.  D.  sought  is  the  same 
number  as  the  G.  C.  D.  of  d  and  a,  which,  as  a  and  b  may  represent  any 
numbers  whatever,  establishes  the  proposition. 

1141.    Application  of  the  foregoing  Princijile. 

(a.)  "When  the  numbers  of  which  the  G.  C.  D.  are  re- 
quired are  such  as  cannot  easily  be  divided  into  factors,  or 
solved  by  the  methods  heretofore  given,  the  principle  just 
demonstrated  may  be  advantageously  applied,  thus  :  — 

(l.)  Divide  the  greater  of  any  two  numbers  by  the  less; 
then  will  the  remainder  obtained  by  this  division  and  the 
smaller  of  the  two  numbers  have  the  same  G.  C.  D.  as  the 
numbers  themselves. 


t 


DnnsoRs  OF  numbkrs.  151 

(c  )  But  as  the  remainder  after  any  division  must  always 
be  less  than  the  divisor,  we  may  divide  the  original  divisor, 
I.  e.,  the  smaller  number,  by  this  remainder.  Now,  if  there 
be  a  remainder  from  this  division,  it  follows,  from  the  propo- 
sition, that  the  G.  C.  D.  sought  must  be  the  G.  C.  D.  of  this 
remainder  and  the  last  divisor ;  and  we  may  divide  the  last 
divisor  by  the  last  remainder. 

(c?.)  But  as  the  remainders  must  constantly  be  decreasing, 
it  follows  that  if  we  continue  this  process,  we  shall  at  last  find 
a  remainder  which  will  exactly  divide  the  preceding,  and  will 
therefore  be  the  G.  C.  D.,  or  we  shall  have  a  remainder  of  1, 
in  which  case  the  numbers  are  prime  to  each  other. 

1.   What  is  the  G.  C.  D.  of  4277  and  9737  ? 

WRITTEN  WORK. 

4277  )  9737  (  2 

1183  )  4277  (  3 

728  )  1183  (  1 

455  )  728  (  1 

273  )  455  (  1 

182  )  273  (  1 

91  )  182  (  2 

00 
Explanation.  —  Dividing  the  greater  number  by  the  less  gives  1183 
for  a  remainder.  Therefore,  the  G.  C.  D.  required  is  the  G.  C.  D.  of 
the  smaller  number,  4277  and  1183.  Dividing  4277  by  1183  gives  728 
for  a  remainder.  Therefore,  the  G.  C.  D.  required  is  the  G.  C.  D.  of 
728  and  1183.  Dividing  1183  by  728  gives  455  for  a  remainder.  There- 
fore, the  G.  C.  D.  required  is  the  G.  C.  D.  of  728  and  1183. 

Proceeding  in  this  way  we  find  that  the  G.  C.  D.  required  is  the  same 
as  the  G.  C.  D.  of  91  and  182,  which  is  91. 

(e.)  The  work  can  frequently  be  much  shortened  by  ob- 
serving that  if  any  remainder  contains  a  prime  factor  which  is 
not  a  factor  of  the  preceding  divisor,  the  factor  may  be  cast 
out  without  affecting  the  G.  C.  D. 

Thus  :  728,  t!ie  second  remainder  in  the  above  example,  is  obviously 
(V  multiple  of  8,  or  2^f  and  2  is  not  a  factor  of  1183,  the  preceding  divi- 


152  MULTIPLES    (»F    NUMBERS. 

sor.  We  may  therefore  cast  out  the  factor  8  from  728.  The  othci 
factor  is  91  ;  th'^vrefore,  the  G.  C.  D.  required  is  the  same  as  the  G.  C.  D. 
of  91  and  1183,  which  is  91. 

The  work,  then,  would  be  written  thus:  — 
4277  )  9737  (2 

1183  )  4277  (  3 

728  =  8  X  91  )  1183  (  13 
273 

00 
Hence,  91  =  G.  C.  D. 

(/.)    What  is  the  greatest  common  divisor  of — 


2.  8383  and  9589  ? 

3.  2021  and  6493  ? 
4     6493  and  8909  ? 


5.  425743  and  584159  ? 

6.  113423  and  836987? 

7.  6941849  and  9128111? 


11^.     Common  Multiples,  Definitions,  and  Properties. 

(a.)  A  multiple  of  a  number  is  any  entire  number  which 
can  be  exactly  divided  by  it. 

(i.)  A  common  multiple  of  two  or  more  numbers  is  a  num- 
ber which  is  a  multiple  of  all  of  them. 

(c.)  The  least  common  multiple  of  two  or  more  numbers 
is  the  smallest  number  which  is  a  multiple  of  all  of  them. 

(d.)  From  these  definitions  and  the  principles  previously 
established,  it  follows  that,  — 

1.  A  multiple  of  a  number  must  contain  all  the  prime  fac- 
tors of  that  number. 

2.  A  common  midtiple  of  two  or  more  numbers  must  con- 
tain all  the  prime  factors  of  each  of  them. 

3.  The  least  common  midtiple  of  two  or  more  numbers  must 
be  the  smallest  number  which  contains  all  the  prime  factors  of 
ectch  of  them. 

116.    Least   Common  Multiple.  —  Method  by  Factors. 
1.    What  is  the  L.  C.  M.*  of  504,  756,  924,  and  1176  ? 

*  L.  C.  M.  means  least  common  muUiple. 


MULTIPLES    OF    NUMHERS.  153 

Solution.  —  The  L.  C.  M.  of  these  numbers  is  the  smallest  number 
which  contains  all  the  prime  factors  of  each  of  them. 
504  =  23  X  32  X  7 
756  =  22  X  33  X  7 
924  =  22  X  3    X  7  X  11 
1176  =  23  X  3    X  72 

The  factors  of  1176  are  23  X  3  X  72  all  of  which  we  take.  The  fac 
tors  of  924  are  22  X  3  X  7  X  11,  all  of  which,  except  11,  we  have 
taken.  We  therefore  introduce  the  factor  11,  which  gives  23  X  3  X  72 
X  11.  The  factors  of  756  are  22  X  33  X  7,  all  of  which,  except  32,  we 
have  taken.  We  therefore  introduce  the  factor  32,  which  gives  23  X  33 
X  72  X  11.  The  factors  of  504  are  23  X  32  X  7,  all  of  which  we  have. 
Therefore,  the  product  of  23  X  33  X  72  X  11,  which  is  116424,  is  the 
L.  C.  M.  required. 

Note.  —  When  the  factors  of  the  L.  C.  M.  have  been  obtained,  much 
labor  in  multiplying  may  be  saved,  by  writing  some  one  of  the  numbers 
in  place  of  its  factors.  Thus,  in  the  example  above,  by  writing  1176 
instead  of  its  factors,  we  should  have  1176  X  32  X  H  =  116424  = 
L.  C.  M.,  as  before. 

What  is  the  least  common  multiple  — 

2.  Of  18,  24,  and  42  ? 

3.  Of  36,  48,  and  60  ? 

4.  Of  132,  144,  and  160? 

5.  Of  98,  126,  and  186? 

6.  Of  364,  637,  and  1547? 

7.  Of  605,  325,  715,  and  1001  ? 

8.  Of  504,  756,  1008,  and  1512  ? 

9.  Of  594,  1386,  and  1782  ? 

10.  Of  735,  945,  1365,  and  2310  ? 

11.  Of  154,231,  264,  and  392? 

117.    Abbreviated  Method. 

(a.)  When  the  factors  of  the  several  numbers  can  readily 
be  determined,  or  after  they  are  found,  much  labor  may  often 
be  avoided  by  considering  what  factors  are  wanted  with  any 
one  of  the  numbers  to  produce  the  L.  C.  M. 

(b.)  Thus,  in  the  example  solved  in  116,  since  the  factors  cannot 
easily  be  recognized,  we  write  them  thus  :  — 


154  MtJLTll'Li:s    OF    NUMBKIJS. 

504  =  23  X  32  X  7 
756  =  22  X  33  X  7 
924  =  22  X  3    X  7  X  11 
1176  =  23  X  3    X  72 
{(  )   Then,  since  1176  contains  all  the  ractors  of  each  of  the  other 
numbers,  except  11  X  32,  the  L.  C.  M.  must  equal  1176  X  11  X  32;  or 
if  divided  by  1176,  the  quotient  would  be  11  X  32,  or  99. 

(d.)  Again.  Since  924  contains  all  the  factors  of  each  of  the  other 
numbers,  except  2  X  7  X  32,  the  L.  C.  M,  must  equal  924  X  2  X  7 
X  32;  or,  divided  by  924,  the  quotient  must  be  2    X  7  X  32,  or  126. 

{e.)  Similar  considerations  will  enable  us  to  obtain  the  L.  C  M.  from 
756  and  from  504.  A  little  practice  will  enable  the  pupil  to  determine 
at  a  glance  from  which  number  the  L.  C.  M.  can  best  be  obtained. 

1.  What  is  the  L.  C.  M.  of  24  and  36  ? 

Solution.  —  Observing  that  36  =  3  X  12,  and  24  =  2  X  12,  it  will 
at  once  be  seen  that  the  L.  C.  M.  may  be  found  by  multiplying  36  by  2, 
or  24  by  3,  and  is  72. 

2.  What  is  the  L.  C.  M.  of  21,  35,  and  56  ? 

Solution.  —  It  Is  obvious  that  56  contains  all  tbe  factors  of  21  and  35, 
except  3  and  5.*     Hence  the  L.  C.  M.  =  56  X  5  X  3  =  840. 

Second  Solution.  —  Since  21  contains  all  the  factors  of  35  and  56, 
except^5  and  8,  the  L.  C.  M.  must  be  21  X  5  X  8  =  21  X  40  =  840. 

Note.  —  In  the  above  example,  it  is  better  to  begin  with  21  than  56, 
because  it  is  easier  to  multiply  21  by  8  X  5,  or  40,  than  to  multiply  56 
by  3  X  5,  or  15. 

3.  What  is  the  L.  C.  M.  of  8  and  9  ? 

Solution.  —  Since  8  and  9  have  no  common  factors,  their  L.  C.  M. 
must  be  their  product,  which  is  72. 

(/•)    What  is  the  least  common  multiple  — 


4.  Of  12  and  25? 

5.  Of  32  and  40  ? 

6.  Of  16  and  27? 

7.  Of  12  and  30? 
.       8.  Of  24  and  35  ? 

9.  Of  42  and  70  ? 

15.  What  is  the  L.  G.  M.  of  15,  18,  80, 12,  and  36  ? 

♦  For  66  =  8  X  7,  35  =  6  X  7,  and  21  =  3  X  7 ;  and  8,  5,  and  3  art 
prime  to  each  other. 


10.  Of  4,  6,  and  12? 

11.  Of  15,  25,  and  35? 

12.  Of  6,  8,  and  9  ? 

13.  Of  4,  9,  and  25  ? 

14.  Of  14,  21,  and  24? 


MULTIPLES    OF    NUMBERS.  155 

Solution.  —  Since  36  is  a  multiple  of  18  and  12,  and  30  is  a  multiple 
of  15,  the  L.  C.  M.  required  must  be  the  L.  C.  M.  of  30  and  36,  which 
equals  5  X  36,  or  6  X  30  =  180. 

{g.)    What  is  the  least  common  multiple  — 

16.  Of  3,  6,  12,  and  18? 

17.  Of  4,  9,  12,  and  18  ? 

18.  Of  5,  3,  6,  and  15  ? 

19.  Of  12,  24,  36,  and  72  ? 

20.  Of  14,  15,  18,  30,  36,  and  42  ? 

21.  Of  9,  10,  11,  12,  and  18? 

22.  Of  98,  154,  198,  and  284? 

23.  Of  72,  108,  180,  and  252  ? 

24.  Of  306,  408,  612,  and  1020  ? 

25.  Of  1548,  2322,  2580,  and  3870  ? 

26.  Of  1872,  4212,  6318,  and  8424  ? 

118.     When  Factors  cannot  easily  he  found. 

(a.)  Since  two  numbers  can  have  no  other  common  factors 
than  those  of  their  greatest  common  divisor,  it  follows  that  the 
least  common  multiple  of  any  two  numbers  may  be  found  by 
multiplying  one  of  them  by  the  quotient  obtained  by  dividing 
the  other  by  their  greatest  common  divisor.  When  prime  to 
each  other,  their  L.  C.  M.  will  be  their  product. 

(6.)  This  principle  may  be  advantageously  applied  when  we 
wish  to  find  the  least  common  multiple  of  numbers  the  factors 
of  which  cannot  easily  be  found. 

1.  What  is  the  L.  C.  M.  of  7379  and  9263  ? 

Solution. — As  these  numbers  cannot  readily  be  divided  into  their 
factors,  we  first  find  their  G.  C.  D.,  which  is  157.  Dividing  7379  by  157 
gives  47  for  a  quotient,  which  must  contain  all  the  factors  of  the  L.C.M. 
that  are  not  found  in  9263.  Hence  the  L.C.M.  must  equal  9263  X  47 
==  435361. 

(c.)  What  is  the  least  common  multiple  — 


2.  Of  5207  and  7493  ? 

3.  Of  2993  and  8651  ? 

4.  Of  3337  and  7471  ? 


5.  Of  3901  and  9047  ? 

6.  Of  6659  and  8083  ? 

7.  Of  7379  and  9263  ? 


Id0  FRACTIONS. 

8.   What  is  the  L.  C.  M.  of  3763,  5183,  and  7261  ? 

Suggestion.  —  First  find  the  L.  C  M.  of  any  two  of  the  numhers,  as 
3763  and  5183,  and  then  the  L.  C.  M.  of  this  result  and  the  remaining 
number. 

(d.)    What  is  the  least  common  multiple  — 
9.    Of  2881,  4171,  and  9313  ? 

10.  Of  2419,  4661,  and  5609? 

11.  Of  3811,  £031,  7519,  and  7661  ? 

12.  Of  8381,  9809,  7361,  and  6817? 


i 


SECTION    X. 

FRACTIONS 

IIO*    Introductory. 

(a.)  If  an  apple,  an  orange,  a  line,  or  any  other  thing,  or  any  num- 
ber, be  divided  into  two  equal  parts,  the  parts  are  called  halves  of  the 
apple.,  orange^  line,  or  of  whatever  may  have  been  thus  divided.  If  two  or 
any  number  of  things  of  the  same  kind  are  each  of  them  divided  into 
two  equal  parts,  the  parts  will,  as  before,  be  called  halves  of  a  thing  of 
that  kind,  and  there  will  be  as  many  times  two  halves  as  there  are 
things  divided. 

(6.)  If  any  thing  should  be  divided  into  t^ree  equal  parts,  the  parts 
would  be  called  thirds  of  the  thing.  If  the  thing  divided  should  be  an 
apple,  the  parts  would  be  thirds  of  an  apple.  If  each  of  several  apples 
should  be  divided  into  three  equal  parts,  all  the  parts,  or  any  portion  of 
them,  would  still  be  called  thirds  of  an  apple.  Again;  if  I  should  cut  ont 
of  an  apple  such  a  part  as  would  be  obtained  by  dividing  it  into  three 
equal  parts,  and  then  cut  from  another  apple  such  another  part,  these 
parts  would  still  be  called  thirds  of  an  apple,  although  one  of  them  came 
from  one  apple  and  one  from  another.  And  so,  however  many  or  few 
such  parts  we  may  take  from  the  same  apple,  or  different  apples,  they 
would  all  be  called  thirds  of  an  apple. 

(c.)  ilcTU'.e,  to  have  a  third  or  thirds  of  any  thing,  qtiantify,  or  number, 
it  is  only  necessary  to  have  one  or  vxore  such  parts  as  would  be  obtained  by 
dividing  the  thing,  quantity,  or  number  into  three  eqiial  parts. 

(d.)  Such  considerations,  extended,  give  the  definitions  of  the  fol 
lowini:  article. 


r 


FKACTIONS.  157 

ISO.    Definitions  of  Halves,  Thirds,  S^c. 

(a.)  Such  parts  as  are  obtained  by  dividing  any  thing  or 
number  into  two  equal  parts,  are  called  halves  of  that  thing 
or  number.  One  such  part  is  called  one  half  of  it ;  two  such 
parts  are  called  two  halves  of  it ;  three  such,  three  halves  of 
it;  &c. 

{h.)  Such  parts  as  are  obtained  by  dividing  any  thing  or 
number  into  three  equal  parts,  are  called  thirds  of  that 
thing  or  number.  One  such  part  is  called  one  third  of  it ;  two 
such,  two  thirds ;  three  such,  three  thirds ;  four  such,  four 
thirds;  &c. 

(c.)  In  like  manner,  such  parts  as  are  obtained  by  dividing 
anything  or  number  into  four  equal  parts,  are  called  fourths 
of  that  thing  or  number  ;  such  as  are  obtained  by  dividing  it 
into  five  equal  parts,  are  called  fifths  ;  into  six,  are  called 
SIXTHS ;  into  seven,  sevenths  ;  into  eight,  eighths  ;  into 
nine,  ninths  ;  into  ten,  tenths  ;  &c 

{d.)    1.  What  are  sevenths  ? 

Answer.  —  Sevenths  of  any  thing  or  number  are  such  parts  as  are 
obtained  by  dividing  it  into  seven  equal  parts. 


2.  What  are  fifths  ? 

3.  What  are  thirds  ? 

4.  What  are  thirteenths  ? 

5.  What  are  fourths  ? 


I 


6.  What  are  ninths  ? 

7.  What  are  tenths  ? 

8.  What  are  halves  ? 

9.  What  are  twenty-firsts  ? 


P 


(e.)  From  the  method  of  obtaining  halves,  thirds,  &c.,  it 
follows  that  — 

Halves  are  equal  parts  of  such  kind  that  two  of  them  would 
equal  a  unit ;  thirds  are  equal  parts  of  such  kind  that  three 
of  them  would  equal  a  unit ;  ^c. 

(f)  Let  the  pupil  now  answer  all  the  preceding  questions 
in  this  article  according  to  the  following  model. 

What  are  sevenths  ? 

Answer.  —  Sevenths  of  any  thing  or  number  are  equal  parts  of  such 
kind  that  it  will  take  seven  of  them  to  equal  that  thing  or  number. 

Note.  —  The  explanations  of  (d.)  and  (/)  are  alike  necessary  to  a 
thorough  understanding  of  fractions,  and  the  student  should  not  rest  sat- 
isfied till  he  has  become  so  familiar  with  both,  as  to  be  able  to  use  cither. 
14 


158  FRACTIONS. 

1^1.    Fractional  Parts. 

(o.)  Such  parts  as  the  above  are  called  fractional 
PARTS,  and  the  arithmetical  expressions  for  them  are  called 
FRACTIONS  ;  hence,  — 

1.  Fractional  parts  of  any  thing,  quantity ,  or  number  are 
such  parts  as  are  obtained  by  dividing  it  into  equal  parts. 

2.  Fractional  parts  of  any  thing,  quantity^  or  number  are 
equal  parts  of  such  hind  that  a  given  number  of  them  will 
equal  a  unit. 

Note.  —  That  from  which  the  fractional  parts  are  obtained  is  always, 
when  considering  the  parts,  regarded  as  a  unit,  and  is  called  the  unit  o' 
the  fraction.     It  may  be,  — 

1.  A  single  object,  as  an  apple,  an  orange. 

2.  A  unit  of  measure,  as  a  foot,  a  yard,  a  bushel. 

3.  The  abstract  unit,  one. 

4.  A  number  of  single  objects  or  units  considered  as  a  collection  or 
whole,  as  6  apples,  1 8  feet,  24  bushels. 

5.  An  abstract  number,  as  5,  8,  12. 

6.  A  fraction,  as  4,  §• 

When  no  particular  unit  is  expressed,  the  abstract  unit  is  the  one 
referred  to. 

(b.)  It  is  obvious  that  the  value  of  any  fractional  part  de- 
pends both  on  the  nature  of  the  unit  and  the  number  of  such 
parts  which  it  takes  to  equal  it. 

Illustrations. —  1.  If  a  large  apple  and  a  small  one  be  each  divided 
into  2,  3,  4,  or  any  other  number  of  equal  parts,  the  parts  of  the  first 
will  be  larger  than  the  corresponding  parts  of  the  second. 

2.  If  several  apples  of  the  same  size  are  divided,  one  into  two  equal 
parts,  another  into  three,  another  into  four,  another  into  five,  &c.,  the 
parts  of  the  first  apple  will  be  larger  than  those  of  any  other,  the  parts 
of  the  second  will  be  smaller  than  tliose  of  the  first,  but  larger  than 
those  of  any  other,  &c. 

(c.)  If  two  equal  units  are  divided,  one  into  two  equal 
parts,  and  the  other  into  four,  each  part  of  the  first  will  be 
equal  to  two  parts  of  the  second ;  if  one  be  divided  into  two 
equal  parts,  and  the  other  into  three  times  as  many,  each  part 
of  the  first  will  be  equal  to  three  parts  of  the  second ;  and, 
generally,  each  part  obtained  by  dividing  a  unit  into  any  num 


FRACTIONS.  l-')9 

ber  of  equal  parts,  will  be  twice  as  large  as  each  would  be  if 
the  unit  had  been  divided  into  twice  as  many  equal  parts ; 
three  times  as  large  as  if  divided  into  three  times  as  many ; 
four  times  as  large  as  if  divided  into  four  times  as  many;  S^c. 

(d,)  In  other  words,  while  the  unit  of  the  fraction  remains 
the  same,  each  half  will  be  twice  as  large  as  each  fourtli,  3 
times  as  large  as  each  sixth,  4  times  as  large  as  each  eighth, 
&:c. ;  each  third  will  be  twice  as  large  as  each  sixth,  3  times 
as  large  as  each  ninth,  4  times  as  large  as  each  twelfth,  &c. ; 
each  fourth  will  be  twice  as  large  as  each  eighth,  3  times  as 
large  as  each  twelfth,  &c. 

(e.)  From  these  considerations,  it  follows  that  in  order 
that  fractional  parts  may  be  of  the  same  denomination,  it  is 
necessary,  — 

1.  That  they  should  be  parts  of  the  same  unit,  or  of  equal 

units ; 

2.  That  they  should  be  obtained  by  dividing  each  unit  into 
the  same  number  of  equal  parts. 

1^^.    Definitions.  —  Method  of  writing  Fractions,  ^c. 

(a.)  A  FRACTION  expresses  the  value  of  such  parts  as  are 
obtained  by  dividing  a  unit  into  equal  parts  ; 

Or,  by  definition  second,  — 

A  FRACTION  expresses  the  value  of  such  equal  parts,  that  a 
certain  number  of  them  will  equal  a  unit. 

(b.)  In  writing  fractions  by  figures,  two  numbers  are  neces- 
sary —  one  to  indicate  the  number  of  parts  into  which  the 
unit  is  divided,  or  (which  is  the  same  thing)  the  number  of 
such  parts  which  it  will  take  to  equal  the  unit ;  the  other  to 
mdicate  how  many  of  the  parts  are  considered. 

(c.)  The  first  of  these  is  called  the  denominator,  because 
it  indicates  the  denomination  of  the  parts.  The  second  is 
called  the  numerator,  because  it  numbers  the  parts. 

(d.)  The  numerator  is  usually  written  above  the  denomi- 
nator, and  separated  from  it  by  a  line. 

Illustrations.  —  Five  sixths  is  written  ^,  5  being  the  numerator,  and  6 
the  denominator. 


160  FRACTIONS, 

Seventh  eighths  is  written  X,  7  being  the  numerator,  and  8  the  denom* 
inator. 

34  twenty -Jirsts  is  written  34  34  being  the  numerator,  and  21  the 
denominator. 

(e.)  Write  the  following  fractions  by  figures,  and  tell  the 
numerator  and  denominator  of  each. 


1. 

3  fourths. 

6. 

1  third. 

2. 

9  twenty-seconds. 

7. 

18  thirteenths. 

3. 

7  halves. 

8. 

5  elevenths. 

4. 

9  seventeenths. 

9. 

15  thirty-firsts. 

5. 

22  ninths. 

(/.)  In  DECIMAL  FRACTIONS  the  numerator  only  is  writ- 
ten, the  denominator  being  determined  by  the  position  of  the 
decimal  point.     (See  33,  a.) 

193.    Exercises  in  explaining  Fractions. 

1.  Explain  the  fraction  f . 

Answer.  —  Five  ninths,  or  five  ninths  of  one,  expresses  the  vahie  of 
five  such  parts  as  would  be  obtained  by  dividing  a  unit  into  nine  equal 
parts.  , 

Another  Form  of  Answer.  —  Five  ninths,  or  five  ninths  of  one,  ex- 
presses the  value  of  five  equal  parts,  such  that  nine  of  them  would  equal 
a  unit. 

2.  Explain  the  fraction  .07. 

First  Form  of  Answer.  —  Seven  hundredths,  or  seven  hundredths  of , 
one,  expresses  the  value  of  seven  such  parts  as  would  be  obtained  by 
dividing  a  unit  into  100  equal  parts. 

Second  Form  of  Answer.  —  Seven  hundredths,  or  seven  hundredths  of 
one,  expresses  the  value  of  seven  equal  parts  of  such  kind  that  100  of 
them  would  equal  a  unit. 

Explain  the  following  fractions  according  to  the  first  form 
of  answer,  and  afterwards  according  to  the  second. 


3. 

f 

6. 

.09. 

9. 

^' 

4. 

H-* 

7. 

.6. 

10. 

.008 

5. 

f^ 

8. 

.27. 

11. 

7^. 

*  25  forty-frstSi  not  25  forty-oneths,  nor  25  forty-ones. 


FRACTIONS. 


161 


12.  M-* 

13.  -yL. 

14.  if.  m 

15.  fft. 


16.  .427. 

17.  .0628. 

18.  .000276. 

19.  tt- 


20.  4f 

21.  f|. 

22.  .0107. 

23.  .002006. 


1341*     Various  Kinds  of  Fractions, 

(a.)  A  simple  fraction  is  one  which  has  but  one  numerator 
and  one  denominator,  each  of  which  is  a  whole  number,  ai  |, 

hi- 

{b.)    Simple  fractions  may  be  proper  or  improper. 

(c.)  A  proper  fraction  is  a  fraction  whose  numerator  is 
less  than  its  denominator,  as  -^-y,  ^^  y^. 

(d.)  An  improper  fraction  is  one  whose  numerator  is  equal 
to,  or  greater  than,  its  denominator,  as  |,  4yS-,  -if^. 

(e.)  A  mixed  number  is  a  whole  number  and  a  fraction,  as 
2^,  which  is  read  two  and  one  fifth  ;  5^,  which  is  read  five 
and  three  sevenths, 

1.  Which  of  the  fractions  in  133  are  proper  ? 

2.  Which  are  improper  ? 

(y.)  A  proper  fraction  is  less  than  1,  because  it  expresses 
less  parts  than  it  takes  to  equal  a  unit. 

{g.)  An  improper  fraction  is  equal  to,  or  greater  than,  1, 
because  it  expresses  as  many  or  more  parts  than  it  takes  to 
equal  a  unit. 

{h.)  An  improper  fraction  is  so  called  because  it  expresses 
a  value,  a  part  or  all  of  which  may  be  expressed  in  whole 
numbers. 

\^5,    Illustrations  of  Operations  on  Fractions, 

Fractions  may  be  added,  subtracted,  multiplied,  and  divided, 
as  whole  numbers  are. 

Thus,  f  +  f  =  f)  just  as  5  days  +  4  days  =  9  days. 
I  -j-  f  =  ^^-,  just  as  7  qts.  +  3  qts.  =  10  qts. 

♦  13  twenty-seconds,  not  13  twenty-twos. 
14* 


I  'V2  FRACTIONS. 

f  i  4-  f  i  =  If.  just  as  23  lbs.  +  21  lbs.  =  44  lbs. 

Again,  f  -f-  8  no  more  equals  f ,  or  f ,  than  3  qts.  -|-  1  pt.  =  4  qts., 
or  4  pt. 

T^  +  ^  no  more  equals  -ff ,  or  ^|,  than  ll§  +  55=16§,or 
16  3. 

Again,  f  —  |-  =  f ,  just  as  $8  —  $3  =  $5,  or  8  lbs.  —  3  lbs.  =■ 
5  lbs. 

ff  —  If  ==  TT^  jnst  as  l.*)  rods  —  13  rods  =  2  rods.* 

f  If-  —  iff  =  iih  jnst  as  358  apples  —  183  apples  =  175  apples. 

Again,  f  —  ^-  no  more  equals  f ,  or  f ,  than  3  pints  —  1  gill  =  2 
pints,  or  2  gills. 

7  —  f  no  more  equals  -fj  or  ^,  than  5  apples  —  4  pears  =  1  apple,  or 
1  pear. 

Again.     5  times  |-  =  \^-,  just  as  5  times  3  apples  =  15  apples. 

8  times  j;^  =  ff ,  just  as  8  times  1 1  oz.  :=  88  oz. 

9  times  iyj|-  =  ^fr^^i  jnst  as  9  times  1387  cubic  inches  =  12483 
cubic  inches. 

Again,  f  are  contained*  3  times  in  f^,  just  as  2  days  are  contained  3 
times  in  6  days. 

5^  are  contained  7  times  in  f  5-,  just  as  3  grains  are  contained  7  times 
in  21  grains. 

Y j^  are  contained  1 8y  times  in  ^f  f ,  just  as  7  apples  are  contained 
18^  times  in  129  apples. 

Again,    i"  of  g^  =  f ,  just  as  ^  of  8  apples  =  2  apples. 

•5-  of  f  ^  =  -jfy  ,  just  as  \  of  24  bushels  =  4^  bushels. 

¥  of  itlf  =  ?T¥T  J  jnst  as  ^  of  1484  miles  =  164f  miles. 

130.    Reduction  of  Whole  or  Mixed  Numbers  to  Improper 
Fractions. 
The  value  of  any  whole  or  mixed  number  may  be  ex- 
pressed by  an  improper  fraction. 

1.  Reduce  8  to  fifths. 

Solution.  —  Since  1  =  f,  8  must  equal  8  times  |-,  or  ^.  Therefore, 
8  =  \^ 

2.  Reduce  9f  to  sevenths. 

Solution.  —  Since  1  =  -f ,  9  must  equal  9  times  7,  or  -*V^,  and  j  added 
are  ^^.    Therefore,  9f  =  ^7^. 

3.  Reduce  4  to  thirds.  |       4.    Reduce  5§  to  thirc's. 


FRACTIONS.  168 


5.  Reduce  6|  to  eighths. 

6.  Reduce  8  to  eighths. 

7.  Reduce  9^  to  fifths. 


5 


8.  Reduce  9f  to  fourths. 

9.  Reduce  439J  to  fourths. 


First  Solution.  —  Since  1=4  fourths,  439  must  equal  439  times  4 
fourths,  which  is  equivalent  to  4  times  439  fourths,  or  1756  fourths,  and 

3  fourths  added  are  1759  fourths.     Therefore,  439f  =  -^-V— • 

Second  Solution. —  Since  there  are  4  fourths  for  each  unit,  there  must 
be  4  times  as  many  fourths  in  any  number  as  there  are  units.  Hence, 
439  =  4  times  439  fourths,  or  1756  fourths,  and  3  fourths  added  are 
1759  fourths.     Hence,  439  J  =  -"¥--• 

Note.  —  Compare  these  reasoning  processes  with  those  given  on  the 
85th  page. 

10.  Reduce  o78f  to  ninths. 

11.  Reduce  14762^  to  sevenths. 

12.  Reduce  74968-py  to  seventeenths. 

13.  Reduce  6483ff  to  forty-sevenths. 

14.  Reduce  438 6^|  to  twenty-eighths. 

lo.    Reduce  427^^f  to  two  hundred  and  seventy-thirds. 
16.    9  =  how  many  times  ^  ?  * 

First  Solution.  —  Since  1  =  J,  9  must  equal  9  times  J,  or  \®-,  and  1 
fourth  is  contained  in  36  fourths  36  times.     Hence,  9  =  36  times  ^. 
Second  Solution.  —  Since  1  =  4  times  5-,  or  f ,  9  must  equal  9  times 

4  times  i,  or  36  times  5-.     Hence,  9  =  36  times  i. 

17.  7  =  how  many  times  ^  ? 

18.  11  =  how  many  times  ^  ? 

19.  49  z=  how  many  times  ^? 

20.  487  =  how  many  times  ^^  ? 

21.  9  =r  how  many  times  ^  ? 

22.  7  z=  how  many  times  .1  ^ 

23.  13  r=  how  many  times  .01  ? 

24.  648  =  how  many  times  .0001  ? 

1)2T.    Reduction  of  Improper  Fractions  to  Whole  or  Mixed 
Numbers. 

The  value  of  any  improper  fraction  may  he  expressed  by  a 
whole  or  mixed  number. 

*  It  will  be  seen  by  the  solutions  that  this  question  is  just  equivalent  to 
'  9  =  how  many  fourths  ?  " 


164  FRACTIONS. 

1.    Reduce  ^g^-  to  ones. 

Since  g  =  1,  ^-  must  equal  as  many  ones  as  there  are  timts  8  in 
43,  which  are  sf  times.     Hence,  -ig^-  =  5f^. 

Note.  —  The  preceding  question  is  precisely  like  this  :  "  43  qts.  =3 
how  many  pecks  ?  "  as  the  following  solution  will  show.     "  Since  8  qtB. 
=  1  pk.,  43  qts.  will  equal  as  many  pecks  as  there  are  times  8  in  43, 
-which  are  5f  times.     Hence,  43  qts.  =  5f  pks." 

8.    Reduce  ^f  to  ones. 


2.  Reduce  ^jp-  to  ones. 

3.  Reduce  \^-  to  ones. 

4.  Reduce  J-f^  to  ones. 

5.  Reduce  ^f^  to  ones. 

6.  Reduce  ^f  |-  to  ones. 

7.  Reduce  fff f  *o  ones. 


9.  Reduce  f  §  to  ones. 

10.  Reduce  -if  ^  to  ones. 

11.  Reduce  ^;f-  to  ones. 

12.  Reduce  ^.i^^r^s.  to  ones. 

13.  Reduce  ^f  ff^^  to  ones. 


128.    Miscellaneous   Questions  involving  Fractions. 

1.  How  much  will  227  yards  of  cloth  cost  at  J  of  a  dollar 
per  yard  ? 

Solution.  —  Since  1  yard  costs  §■  of  a  dollar,  227  yards  will  cost  227 
times  f  of  a  dollar,  equivalent  to  3  times  ^f  ^  of  a  dollar,  which,  found 
by  multiplying  227  by  3,  is  ^f  ^  of  a  dollar,  equal  (by  12'S')  to  170;^ 
dollars. 

Note.  —  Compare  the  preceding  example  and  solution  with  the  fol- 
lowing :  — 

How  many  bushels  of  grain  in  227  bags,  each  holding  3  pecks  ? 

Solution.  —  Since  1  bag  holds  3  pecks,  227  bags  will  hold  227  times 
3  pecks,  equivalent  to  3  times  227  pecks,  which,  found  by  multiplying 
227  by  3,  is  681  pecks,  equal  to  170  bushels,  I  peck. 

2.  What  will  493  lbs.  of  tea  cost  at  f  of  a  dollar  per  lb.  ? 

3.  What  will  1286  bu.  of  apples  cost  at  f^  of  a  dollar 
per  bu.  ? 

4.  What  will  347  gal.  of  burning  fluid  cost  at  ^^  of  a  dol- 
lar per  gal.  ? 

5.  How  many  acres  in  169  house  lots,  each  containing  f^- 
»f  an  acre  ?  \ 

6.  How  many  acres  in  3  lots,  each  containing  t§f  |^  acres  ? 

7.  How  many  quarts  in  9  boxes,  each  holding  |f  of  8 
quart  ? 


FKACTIONS.  165 

8.  How  many  pounds  of  silver  in  28  bars,  each  weighing 
f  f  ^^  of  a  pound  ? 

9.  How  many  tons  of  hay  in  37  loads,  each  containing 
i^  of  a  ton  ? 

10.  How  much  will  7  acres   of  land  cost  at  $275f  per 

acre? 

Note.  —  Compare  the  last  example  with  this  :  "  How  much  corn  in 
7  bins,  each  holding  275  bushels,  3  pecks  ?  " 

11.  How  much  will  6  acres  of  land  cost  at  $493|^  per 
acre  ? 

12.  If  a  ship  sails  uniformly  at  the  rate  of  179^^  miles 
per  day,  how  far  will  she  sail  in  5  days  ? 

13.  If  a  steamship  sails  uniformly  at  the  rate  of  4179f|^ 
rods  per  hour,  how  many  rods  will  she  sail  in  85  hours  ? 

14.  How  many  baskets,  each  holding  |^  of  a  pe/»k,  can  be 
fdled  from  82f  pecks  of  corn  ? 

Solution. —  Since  |-  of  a  peck  will  fill  one  basket,  82 1  pecks  will  fill 
as  many  baskets  as  there  are  times  |-  in  82f .  But  82f  =  ^f-,  which 
contains  ^  as  many  times  as  659  contains  7,  which  is  94^  times.  Hence, 
947"  baskets  can  be  filled. 

Note.  —  Compare  the  above  question  and  solution  with  the  follow- 
ing :  "  How  many  baskets,  each  holding  7  quarts,  can  be  filled  f-om 
82  pk.  3  qt.  of  corn  1 " 

Solution.  —  Since  7  quarts  will  fill  one  basket,  82  pk.  3  qt.  will  fill  as 
many  baskets  as  there  are  times  7  qt.  in  82  pk.  3  qt.  82  pk.  3  qt.  =  659 
quarts,  which  contains  7  quarts  as  many  times  as  659  contains  7,  which 
is  94y  times.     Hence,  947  baskets  can  be  filled. 

15.  How  many  vest  patterns,  each  containing  f  of  a  yard, 
can  be  cut  from  a  piece  of  vesting  containing  25^  yards  ? 

16.  How  many  books,  at  f  of  a  dollar  a  piece,  can  be 
bought  for  232^  dollars  ? 

17.  How  many  baskets,  each  holding  ^|-  of  a  bushel,  can 
be  filled  from  4537^2-  bushels  of  apples  ? 

18.  How  many  barrels,  holding  2^  bushels  each,  can  be 
filled  from  59f  bushels  of  apples  ? 

Note.  —  Compare  the  last  question  with,  —  "  How  many  jugs,  hold- 
up 2  gal.  1  qt.  each,  can  be  filled  from  59  gal.  3  qt.  of  fiuid  ?  " 


166  FRACTIONS. 

19.  How  many  spoons,  each  weighing  7f  dwt.,  can  be 
made  from  328|^  dwt.  of  silver  ? 

20.  How  many  yards  of  cloth,  at  2^f  dollars  per  yard,  can 
be  bought  for  89|^  dollars  ? 

21.  How  many  days,  at  $2^^  per  day,  must  a  man  laboi 
to  earn  $237^f  ? 

22.  A  man,  who  owned  247^  dwt.  of  silver,  had  it  made 
into  spoons,  each  weighing  4^^  dwt.  How  many  spoons  did 
it  make  ? 

23.  What  is  the  sum  of  |^  +  -f^  +  ^^2"  +  T^  +  i^  + 

Note.  —  Compare  the  above  example  with  1 1  in.  -|-  7  in.  +  9  in-  -j- 
4  in.  +  7  in.  +  5  in.  =  43  in.  =  3  ft.  7  in. 

What  is  the  sum  — 

24.  Ofi  +  l  +  i  +  f  +  i? 

25.  Of  -rV  +  -  +  ,-V  +  ,9^  +  ^  +  1^  +  ft  ? 

26.  Of  If  +  19  +  2^  +  1^  +  f^  +  /^  +  ^1? 

27.  Of  13f  +  13|  +  2f  +  22f  +  43f  ? 

NoTK.  —  Compare  the  27th  example  with  the  following:  — 13  w 
3  da.  4*  13  w.  5  da.  +  2  w.  4  da.  +  22  w.  6  da.  -f-  ^3  w.  2  da.  =  95  w. 
6  da.  To  make  the  resemblance  to  compound  addition  still  more  ap- 
parent, we  give  two  forms  of  writing  the  work  of  the  27th  example,  and 
place  opposite  them  the  work  of  the  example  in  this  note. 

Ones.    Sevenths.  w.        da. 

13^  =  13         3 

13f  =13  5 

2f  =    2  4 

22f  =  22  6 

43f  =  43  2 

95f  =  95        6  =  Ans.  95        6  =  Am, 

What  is  the  value  — 

28.  Of  n^Sj.  +  18t^  +  23^T  +  42t^  +  27^^? 

29.  Of  24^  +  96t  +  14t  +  26^  +  55|  ? 

\ 

*  These,  like  the  compound  numbers  of  56,  can  beat  be  reduced  u 
they  are  added ;  thus,  y^  +  rV  (by  adding  one  of  the  -^^  to  ^^)  =  1-j^  J 
and  l-j^  ■\-  -^^  (by  adding  3  of  the  -j^  to  the  ■^)  =  2^^ ;  &c. 


13 

5 

2 

4 

22 

6 

43 

2 

i 


FRACTIONS. 


167 


30. 
31. 
32. 
33. 
34. 


Of  237^^  +  496^^  +  849/^  +  591^§  +  438^^  ? 
Of  l-f? 


Of    12  8 
^^    TFT 


rgy 


9   ? 


8325J_ 
:6S 


4  9468    ? 
2  6"3-F¥3-  • 


Of  24/^  —  19|i  ? 

Note.  —  Compare  the  34th  with  the  following :  —  What  is  the  ralue 
of  24  qr.  71b.  —  19qr.  21  lb. 

By  writing  the  work  of  the  34th  example  in  the  following  form,  and 
placing  opposite  it  the  work  of  the  example  in  this  note,  the  resem- 
blance of  the  two  is  very  clearly  exhibited. 

Ones.    Twenty-fifths.  Qr.        lb. 

24^V  =  24         7  24         7 

19f^  =  19       21  19       21 


Ans. 


4ii=    4       11 

What  is  the  value  — 

35.  Of  247ff— 159^1? 

36.  Of  4327f f  —  2589||  ? 

37.  Of  673211^1 -4694fH^? 

38.  Of  479^117 -214iH^? 
.39.   Of  284i|§f|  — 283if|f|? 

40.    Of  9243f|ei~3842f|ef? 


11  =»  Ans. 


1Q9.     One  Number  a  Fractional  Part  of  another^ 

What  part  of  4  is  1  ? 

Answer.      1  is  one  fourth  of  4,  because  4  times  1  are  4.* 

What  part  — 

1.  Of  6  is  1  ?  3.   Of  9  is  1  ?  5.   Of  16  is  1  ? 

2.  Of  2  is  1  ?  4.   Of  12  is  1  ?  6.   Of  243  is  1  ? 
7.    What  part  of  8  is  3  ? 

Solution.  —  Since  1  =  i|  of  8,  3  must  equal  f  of  8. 
What  part  — 


8.  Of  9  is  4  ? 

9.  Of  11  is  7  ? 


10.  Of  3  is  2  ? 

11.  Of  8  is  5? 


12.  Of  6  is  5? 

13.  Of  13  is  4i 


} 


i.  e.,  Because  1  taken  4  times  will  equal  4 


168 


FRACTIONS. 


14.    What  part  of  7  is  9  ? 

Solution.  —  Since  1  =  4-  of  7,  9  must  equal  ^  of  7,  or  if  times  7. 

"What  part  — 


15.  Of  9  is  13  ? 

16.  Of  10  is  87? 

17.  Of  19  is  39  ? 


18.  Of  47  is  63? 

19.  Of  39  is  19? 

20.  Of  413  is  527  ? 

21.  What  part  of  8  yards  is  1  yard  ? 

First  Form  of  Answer.  1  yard  =  ^  of  8  yards,  because  8  times  1 
yard  =  8  yards. 

Second  Form  of  Answer.  1  yard  is  the  same  part  of  8  yards  that  1  if 
of  8,  which  is  ^.    Hence,  1  yard  =  |-  of  8  yards. 

What  part  — 

22.  Of  9  ft.  is  1  ft.  ? 

23.  Of  4  m.  is  3  m.  ? 

24.  Of  4  lb.  is  3  lb.  ? 
28.    What  part  of  f-  is  f  ? 

Answer,    f  is  the  same  part  of  ^  that  3  is  of  4,  which  is  f . 
^^  is  f  of  t. 

What  part  — 


25.  Of  4  cwt.  is  3  cwt  ? 

26.  Of  10  da.  is  7  da.? 

27.  Of  8  yr.  is  5  yr.  ? 


Hence^ 


29.  Of  I  is  f? 

30.  Of^t^is-rf^? 

31.  Of  il^  is  if? 

32.  Of^VVis^V^? 


33.  Of-Z^is^^? 

34.  Of^isT^? 

35.  Off  is  I? 

36.  Of  ft  is  it? 


37.   What  part  of  1  bushel  is  3  pecks  ? 

Solution.     1  bu.  =  4  pk.,  and  3  pk.  =  f  of  4  pk.     Hence,  3  pk.  ^ 


i  of  1  bu. 

What  part  — 

38.  Of  1  m.  is  5  fur.  ? 

39.  Of  lib.  is  7  i? 

40.  Of  1  A.  is  3  R.? 


41.  Of  1  yd.  is  2  ft.  ? 

42.  Of  1  ft.  is  1  in.  ? 

43.  Of  1  T.  is  13  cwt.  ? 


44.   What  part  of  3  da.  is  1  w.  ? 

Solution.     1  w.  =  7  da.,  and  7  da.  =  |-  of  3  da.,  or  2^  times  3  da. 

What  part  — 


45.  Of  5  d.  is  1  s.  ? 

46.  Of  7  8.  is  £1  ? 


47.  Of  13  dr.  is  1  oz.  ? 

48.  Of  9  dwt.  is  1  oz.  ? 


FRACTIONS. 


loy 


I 


49.   Of  11  i  is  1  lb.  ?  I      50.    Of  8  da.  is  1  w.  ? 

51.   What  part  of  |  is  1  ? 

Solution.     1  =  f ,  and  f  is  the  same  part  of  f  that  6  is  of  5,  which  la 
Hence,  1  =  f  of  f ,  or  1-|^  times  -g-. 


57.  Of  V^  is  1  ? 

58.  Of  2/  is  1  ? 

59.  Of  ^f  is  1  ? 

60.  Of  ^\\  is  1  ? 

61.  OfyM^isl? 


What  part  — 

52.  Of  A  is  1  ? 

53.  Of  f  is  1  ? 

54.  Of  ^  is  1  ? 

55.  Of  ^%s  is  1  ? 

56.  Of  /j  is  1  ? 

62.  What  part  of  If  is  2f 

Solution,     if  =  -V- ;  2f  =  -V- ;  and  V  =  f T  of  V  =  if?  times 

63.  What  part  of  1  m.  3  fur.  is  2  m.  5  fur.  ? 

Solution.     1  m.  3  fur.  =  11  fur. ;  and  2  m.  5  fur.  =  21  fur. ;  21  fur.  = 
fi  of  11  fur.,  or  if^  times  11  fur. 

What  part  — 

64.    Of  Id.  3qr.  is  lid.  Iqr.? 


Of  if  is  lip 
Of  8  w.  3  da.  is  2  w.  4  da.  ? 
Of  Sf  is  2|-  ? 

Of  5  da.  7  h.  is  3  da.  11  h.? 
..,   Of  5jVis3^i? 

70.  Of  11  h.  27  m.  is  21  h.  38  m.  ? 

71.  Of  15  pk.  is  3  pk.  7  qt.  ? 

72.    What  part  of  1  lb.  is  3  oz.  5  dwt..  17f  gr.  ? 

Reduce  1  lb.  and  also  3  oz.  5  dwt.  17|gr.  to  the  lowe8(^ 


65. 
66, 

67. 
68. 
69, 


Suggestion.       .  „. 

denomination  mentioned,  i.  e.,  to  fifths  of  a  grain 


What  part  — 

73.  Of  1  gal.  is  3  qt.  1  pt.  3  gi.  ? 

74.  Of  £1  is  17s.  lid.  2qr.? 

75.  Of  1  w.  is  5  da.  11  h.  35  m.  29|  sec.  ? 

76.  Of  1  T.  is  13  cwt.  2  qr.  19  lb.  13  oz.  11^  dr.  ? 

77.  Of  4  yd.  2  ft.  7  in.  is  1  yd.  1  ft.  11  in.  ? 

78.  Of  17  A.  3  R.  16  sq.  rd.  is  2  A.  1  R.  28  sq.  rd.  ? 

15 


170 


FRACTIONS. 


130.     Other  Methods  of  expressing  the  Value  of  a  Fraction, 
(«.)    1.  Explain  the  fraction  f . 


f  =  3  times  j-,  as  f  of  a  number 


=  3 

times  \  of  the 

8. 

9. 

10. 

Of  .4 
Of  .007 
Of  .0346 

Answer. 
number. 

In  the  same  way  explain  the  meaning  — 

2.  Off.  5.  om. 

3.  Of*.  6.   OfyVV 

4.  Of  f.  7.   Of  .15 
(h.)   Explain  each  of  the  above  by  the  following  method, 
f^,  or  |-  of  1,  =  i  of  3,  as  f  of  a  number  =  :|-  of  3  times  the  num 

ber.    For  if  each  of  3  equal  things  should  be  divided  into  4  equal  parts, 
and  1  of  the  parts  of  each  taken,  the  result  would  equal  f  of  1. 

Note.  —  This  may  be  illustrated  to  the  eye,  by  taking  3  equal  lines, 
=  and  dividing  them  into  4  equal  parts,  arranged  thus 

^==- =rrz=  one  of  the  parts  will  then  contain  ^  of  3  lines,  which, 

as  will  be  seen,  is  equal  to  f  of  a  line. 

11.    To  what  simple  fraction  is  J  of  3  equal  ? 
Answer.    :^  of  3  =  f  of  1  =  f . 


(c.)    To  what   simple   fraction   is 

each   of  the  following 

uai  r 

12.  ^  of  2. 

13.  1  of  4. 

14.  i^i  of  6. 

15.  ^  of  7. 

16.  ^  of  11. 

17.  V^of  4. 

18.  T^j  of  15. 

19.  •^"'of49. 

20.  TVof832. 

21.  1  of  98. 

22.  2V  of  493. 

23.  jVof861. 

131.    To  find  a  Fractional  Part  of  a  Number, 

1.   What  is  I  of  5459  ? 

First  Solution.  |  of  5459  =  7  times  ^  of  5459 ;  i  of  5459,  found 
by  dividing  by  8,  is  682f ,  and  7  times  682t  =  4776|-. 

/Second  Solution.  ^  of  5459  =  -^  of  7  times  5459  ;  7  times  5459  ==■ 
38213,  and  ^  of  38213  =  4776^. 

WRITTEN   WORK. 

By  First  Solution.  By  Second  Solution. 

8  )  5459 

.  7 
4770^  =  A'is. 


5459 


8  ^  38213 


4776|  =  An$. 


FRACTIONS.  171 

The  following  forms  of  writing  the  work  exhibit  very  cleany  the 
reason  for  every  step  which  is  taken.  The  letters  a,  b,  &c.,  are  used  as 
explained  in  the  note  on  the  88th  page. 

First  Form.  Second  Form. 

9  —  5459  a  =    5459 

^  of  a  =  b  =    682t  7  X  a  =  b  =  38213 

7Xb  =  ^ofa  =  4776f  i  of  b  =  f  of  a  =    4776f 

Note.  —  It  will  be  seen  that  by  both  methods  of  solution,  we  multi- 
ply by  the  numerator,  7,  and  divide  by  the  denominator,  8.  The  only 
difference  between  them  is,  tliat  by  the  first  we  divide  before  we  midtiply, 
while  by  the  second  we  multiply  befare  we  divide.  The  first  method 
in^o^ves  somewhat  smaller  numbers  than  the  second. 

What  is  the  value  of  each  of  the  following  ? 

2.  f  of  3843.       I     4.   f  of  7148.       i      6.  ^  of  46935% 

3.  I- of  7987.       [     5.  -1  j  of  58437.    I     7.  |fJof87516. 
8.   What  is  the  value  of  .0014  of  16.427  ? 

First  Solution.  .0014  of  16.427  =  14  times  .0001  of  16.427  ;  .0001 
of  16.427,  found  by  removing  the  point  four  places  towards  the  left,  is 
0016427,  and  14  times  this  is  .0229978  =  Ans. 

The  work  may  be  written  in  full  after  either  of  the  following  models 

First  Model. 

a  =  16.427 
.0001  of  a  =  b  =      .0016427 
14  times  b  =  .0014  of  a  =      .0229978  =  Ans, 

Second  Model. 
10000  )  16.427 


.0016427 
14 


.0229978  =  Ans. 

Second  Solution.  .0014  of  16.427  =  .0001  of  14  times  16.427;  14 
times  16.427  =  229.978,  and  .0001  of  229.978,  found  by  removing  the 
point  four  places  towards  the  left,  is  .0229978  =  Ans. 

The  work  may  be  written  in  full  after  either  of  the  following  models 

First  Model. 

a=    16.427 
14  times  a  =  b  =  229.978 
.0001  of  b  =  .0014  of  a  =        .0229978 


172  FRACTIONS. 

Second  Model. 
16.427 
14 


10000  )  229.978 


.0229978 
Note.  —  It  will  be  seen  that  by  both  solutions,  we  multiply  by  the 
numerator^  14,  and  divide  by  the  denominator,  10000;  i.e.,  we  multiply 
by  14,  and  remove  the  point  four  places  towards  the  left.  The  only  dif- 
ference between  them  is,  that  in  the  first  we  remove  the  point  before  we 
multiply,  while  in  the  second  we  remove  it  afterwards.  Some  of  the  above 
written  work  might  have  been  avoided  by  raulti})lying  by  14,  and  chang- 
ing the  position  of  the  point  at  the  time  of  writing  the  product,  thus  :  — 

16.427 
.0014 


.0229978 

In  like  manner  we  may  get  .07  of  a  number  by  multiplying  by  7, 
and  removing  the  point  two  places  towards  the  left ;  we  may  get  5.2  *  of  a 
number  by  multiplying  by  52,  and  removing  the  point  one  place  towards 
the  left ;  &c. 

What  is  the  value  of  each  of  the  followinof  ? 


9. 

.03  of  2589 

14. 

.006  of  .006 

10. 

.005  of  3.7479 

15. 

2.5  of  2.5 

11. 

.28  of  437.96 

16. 

.372  of  5.46 

12. 

4.2  of  67.93 1 

17. 

.0004  of  200 

13. 

.25  of  .25 

18. 

.006  of  .006 

133*    To  multiply  hy  a  Fraction. 

1.   What  is  the  product  of  .865  multiplied  by  f  ? 

Suggestion.  —  The  product  of  865  multiplied  by  ^  equals  ^  times 
865,  which  is  the  same  as  ^  of  865,  and  may  be  found  by  methods  be- 
fore explained. 

Note.  —  This  follows  from  the  definition  of  multiplication  —  A 
process  by  which  we  ascertain  how  much  any  given  number  will  amount  [to, 
if  taken  as  many  times  as  there  are  units  in  some  other  given  number.  In 
multiplying  by  more  than  a  unit,  then,  more  than  once  the  multiplicand 
is  taken,  while  in  multiplying  by  less  than  a  unit,  less  than  once  the  mul- 

♦  It  will  be  more  convenient,  for  purposes  of  explanation,  to  regard  this 
as  an  improper  fraction  than  as  a  mixed  number. 

t  The  iminber  of  which  a  fractional  part  is  taken  may  be  regarded  either 

afi  a  mixed  nuinluT Or  an  improper  fraction. 


FRACTIONS.  173 

liplicand  (i.  e.,  some  fractional  part  of  it)  is  taken  ;  the  product  being 
always  the  same  part  of  the  multiplicand  that  the  multiplier  is  of  unity. 
When,  therefore,  the  multiplier  is  greater  than  unity,  the  product  will 
be  greater  than  the  multiplicand;  and  when  the  multiplier  is  less  than 
unity,  the  product  will  be  less  than  the  multiplicand.  In  multiplying 
^y  i".  ?  of  once  the  multiplicand  is  taken ;  in  multiplying  by  f,  f  of 
once  the  multiplicand  is  taken  ;  &c. 

This  conclusion  is  in  exact  accordance  with  the  principle,  that  the 
larger  the  multiplier  the  larger  the  product,  and  the  smaller  the  multi- 
plier the  smaller  the  product,  so  long  as  the  multiplicand  remains  the 
same  ;  or,  to  speak  more  definitely,  that  the  product  of  a  number  mul- 
tiplied by  ^  of  another,  is  equal  to  3-  of  its  product  multiplied  by  the 
whole  of  the  other ;  that  the  product  of  a  number  multiplied  by  |-  of 
another,  is  equal  to  f  of  its  product  multiplied  by  the  whole  of  the 
other;  &c. 

Thus,  12  multiplied  by  6  =  72. 

12  multiplied  by  ^  of  6,  or  3,  =  36,  which  is  ^  of  72. 
12  multiplied  by  ^  of  6,  or  2,  =  24,  which  is  i  of  72. 
12  multiplied  by  f  of  6,  or  4,  =  48,  which  is  |  of  72 ;  &c 
So,  since  12  multiplied  by  1  =  12, 

12  multiplied  by  2-  of  1,  or  ^,  must  equal  2"  of  12,  or  6. 
12  multiplied  by  ^  of  1,  or  ^,  must  equal  ^  of  12,  or  4. 
12  multiplied  by  f  of  1,  or  §,  must  equal  f  of  12,  or  8;  &c. 
Note.  —  By  the  foregoing  explanations,  it  appears  that  the  change 
of  denomination  mentioned  in  the  note  under  74,  {h.)  is  caused  by  the 
fact  that  to  multiply  by  a  fraction  requires  a  division  as  well  as  a  mul- 
tiplication. 

What  is  the  product  — 


5.  Of  ^  times  6964? 

6.  Of  .7  times  6397  ? 

7.  Of  .06  times  59.37  ? 


2.  Of  f  times  8649  ? 

3.  Of  f  times  6743  ? 

4.  Of -^  times  16872? 
8.   What  is  the  product  of  3f  times  2948  ? 
Suggestion.    3|  times  2948  =  8  times  2948  -f-  f  of  2948. 

WBITTESr  WORK. 

First  Form. 
a  =    2948  9  )  2948  =  a 

3^  

327|  =  ^  of  a 

5 

1637^  =  f  of  a 


S  X  a  =  b  =    8844  5 

fofa  =  c=    I637|- 


b4  c«=3^4-a=  10481^  =  An$. 
15  ♦ 


174  FRACTIONS. 

Second  Form. 

a=  2948 


3  X  a  =  b  =  8844 

^  of  a  =  c  =  327f 

4Xc  =  d=    1310|  =  |ofa 

»*b  +  c4-d  =  3f  +  a=  10481^  =  Ans. 

Note.  —  The  second  of  the  above  forms  is  preferable  to  the  first, 
inasmuch  as  it  presents  to  the  eye,  in  a  compact  and  convenient  form, 
all  the  numbers  which  it  is  necessary  to  use,  and  avoids  all  side  work. 
The  references  by  means  of  letters  may  be  omitted  as  soon  as  the  pro- 
cesses and  explanations  are  familiar. 


What  is  the  product  — 
9.    Of  8^  times  6794? 

10.  Of  26f  times  2579? 

11.  Of  43^  times  45687  ? 


12.  Of  3|- times  85476  ? 

13.  Of  25t4j.  times  2764? 

14.  Of  698J  times  29679  ? 


133.    Practical  Problems. 

1.  How  much  will  |  of  an  acre  of  land  cost  at  $478.36 
per  acre  ? 

Solution.  —  If  1  acre  costs  $478.36,  ^  of  an  acre  will  cost  |-  of 
$478.36,  which,  found  in  the  usual  way,  is  $418.56^. 

First  Form.  Second  Form. 

.     8  )  $478.36  a  =    $478.36    =  cost  of  1  acre. 

^  of  a  =  b  =  $  59.79^  =  cost  of  ^  of  an  A. 

$59.79 J  7  X  b  =  $418.56|  =  cost  of  §  of  an  A. 
7 


$418.56|  =  Ans. 

Note.  —  The  second  form  is  better  than  the  first,  when  a  full  state- 
ment and  explanation  of  the  successive  steps  are  required  ;  but  in  other 
cases  the  first  is  preferable. 

2.  How  much  will  .9  of  an  acre  of  land  cost  at  $394.26 
per  acre  ? 

»  Since  ^  =  7  +  J. 


FRACTIONS.  175 

Solution.  —  If  1   acre   costs   S394.26,  .9  of  an   acre  will  cost  .9  of 
.   $394.26,  which,  found  as  before  explained,  is  $354,834.     The  following 
exhibits  the  written  work. 

Full  Form.  Abbreviated  Form. 

a  =  $394.26    =  cost  of  1  acre.  $394.26 
.1  of  a  =  b  =  $  39.426  =  cost  of  .1  of  an  A.  .9 

9  X  b  =  $354,834  =  cost  of  .9  of  an  A.  

f  $354,834  =  Ans. 

S.  How  much  is  -f  of  a  vessel  ^Yortll,  if  the  whole  vessel  is 
worth  $29951.88? 

4.  I  bought  f  of  an  acre  of  land  at  $397.36  per  acre. 
How  much  ought  I  to  pay  for  it  ? 

5.  If  a  bushel  of  tomatoes  weighs  48.75  lb.,  what  will  ^^ 
of  a  bushel  weigh  ? 

6.  What  will  .4  of  a  ton  of  iron  cost  at  $47.93  per  ton  ? 

7.  What  will  .8  of  a  bag  of  coffee  cost  at  $27.94  per  bag? 

8.  Mr.  Foster  bought  4377  bushels  of  corn,  and  sold  .37 
of  it  to  Mr.  Gardiner,  .43  of  it  to  Mr.  Calder,  and  the  rest  to 
Mr.  Godfrey.     How  many  bushels  did  he  sell  to  each  ? 

9.  What  will  9|^  acres  of  land  cost  at  $126  per  acre  ? 

10.  Wliat  will  32ff  acres  of  land  cost  at  $139.50  per 
acre  ? 

11.  The  ship  Surprise  is  valued  at  $25427.  Joseph  Ward 
owns  §  of  her,  and  Samuel  Lowe  owns  the  remainder.  What 
is  the  value  of  Mr.  Ward's  share  ?     Of  Mr.  Lowe's  ? 

12.  The  brig  Maria  sailed  from  Philadelphia  to  Boston, 
with  207  T.  13  cwt.  2  qr.  19  lb.  of  coal,  f  of  which  was  landed 
at  one  wharf,  and  the  remainder  at  another.  What  was  the 
weight  of  that  landed  at  each  wharf? 

13.  If  John  walks  ^  as  fast  as  George,  how  far  w^ill  John 
walk  while  George  is  walking  97  m.  6  fur.  37  rd.  2  yd.  ? 

14.  A  company  of  9  California  gold  diggers  found  in  1 
month  37  lb.  4oz.  16dwt.  17  gr.  of -gold,  which  they  divided 
equally.     What  was  the  share  of  each  ? 

15.  An  English  ship,  valued  at  £8743,  carries  a  cargo 
which  is  worth  ^  as  much  as  the  ship.  How  many  £,  s.,  and 
4.  is  the  cargo  worth  ?     In  a  storm  the  sailors  were  obli^d 


176  FRACTIONS. 

to  throw  I  of  the  cargo  overboard.     What  was  the  value  of 
what  they  threw  overboard  ? 

16.  A  merchant  bought  17  T.  14  cwt.  2  qr.  23  lb.  10  oz.  of 
sugar.  He  sold  f  of  it  to  one  man,  f  of  the  remainder  to 
another,  and  what  there  was  left  to  another.  What  was  the 
weight  of  that  which  he  sold  to  the  first  man  ?  To  the 
second  ?     To  the  third  ? 

Suggestion.  —  After  selling  ^  of  the  sugar,  he  must  have  had  ^  of  it 
left,  and  -f  of  this,  or  what  he  sold  to  the  second  man,  must  be  ^  of  f » 
or  T"  of  the  original  quantity ;  and  the  remaining  -f  of  ^,  or  y  of  the 
original  quantity,  must  have  been  sold  to  the  third  man. 

Or,  the  share  of  the  third  man  may  be  determined  thus  :  Since  the 
first  man  took  y  of  the  sugar,  and  the  second  ^,  both  took  ^  of  it,  which 
would  leave  y  for  the  third  man. 

17.  A  father,  dying,  left  to  his  son  an  estate  valued  at 
$87646.44.  The  first  year  after  the  son  came  into  possession 
of  the  property,  he  spent  ^  of  it  in  dissipation,  the  second  year 
he  spent  f  of  the  remainder,  and  at  the  end  of  the  third  year 
he  was  penniless.  How  much  did  he  spend  in  the  first  year  ? 
How  much  in  the  second  year  ?     How  much  in  the  third  ? 

Suggestion.  —  Compare  this  example  with  the  last 

18.  If  Mr.  Smith  uses  9.678  tons  of  coal  per  year,  and 
Mr.  Parkhurst  uses  .7  as  much,  how  many  tons  does  Mr. 
Parkhurst  use  ? 

Note.  —  The  term  per  cent  is  often  used  in  arithmetic,  and  in  busi- 
ness transactions,  instead  of  one-hundredths.  Thus,  6  per  cent  has  the 
same  meaning  as  6  one-hundredths.     8  per  cent  =  .08  =  t^tj-. 

5  per  cent  of  340  acres  =  .05  of  340  acres. 

6  per  cent  of  $86.38  =  .06  of  $86.38. 
Every  number  is  100  per  cent  of  itself. 

19.  A  teamster  carted  7486  bushels  of  potatoes  to  a  rail- 
road depot,  receiving  in  payment  9  per  cent  of  them.  How 
many  bushels  did  he  receive  ? 

20.  A  man  who  owned  378.37  acres  of  land  gave  8  per 
cent  of  it  to  his  son.  How  many  acres  did  he  give  to  hia 
son? 


FRACTIONS.  177 

21.  A  city  merchant  agrees  to  sell  for  a  countrj  farmer 
whatever  produce  the  latter  may  send  to  him,  on  condition 
that  he  shall  receive  for  his  trouble  5  per  cent  of  the  money 
he  receives  for  the  produce.  Under  this  arrangement  he  sells 
produce  to  the  value  of  $17G8.37.  What  ought  he  to  receive 
for  his  trouble  ? 

Note.  —  A  merchant  who  makes  it  his  business  to  buy  and  sell 
goods  for  others  is  called  a  Commission  Merchant.  The  money  he 
receives  for  his  services  is  called  his  Commission.  For  instance  :  In 
the  last  example,  the  merchant  receives  a  commission  of  5  per  cent  on 
the  value  of  the  produce  he  sells.  He  will  deduct  this  commission  from 
the  amount  of  the  sales  before  paying  the  farmer. 

22.  A  commission  merchant  sold  cloth  for  a  manufacturer 
to  the  amount  of  $7643.79,  receiving  a  commission  of  3  per 
cent.  What  did  his  commission  amount  to  ?  How  much 
money  ought  he  to  pay  the  manufacturer  ? 

23.  A  commission  merchant  buys  goods  for  me  to  the 
amount  of  $387.46,  for  which  he  charges  a  commission  of  2 
per  cent.  What  will  his  commission  amount  to  ?  How  much 
money  must  I  send  him  to  pay  for  the  goods  and  commission  ? 

24.  Mr.  Moore  borrowed  some  money  of  Mr.  Boyden, 
agreeing  to  pay  him,  for  each  year's  use  of  it,  a  sum  equal  to 
6  per  cent  of  the  money  he  had  borrowed.  He  borrowed 
$125.63,  and  kept  it  just  one  year.  How  much  ought  he  to 
pay  for  the  use  of  it  ?  How  much,  then,  ought  he  to  pay  Mr. 
Boyden  in  all  ? 

Note.  —  Money  paid,  as  in  the  above  example,  for  the  use  of  money, 
is  called  Interest.  The  money  used  is  called  the  Principal.  Inter- 
est is  usually  reckoned  at  a  certain  per  cent  of  the  principal  for  each 
year  that  it  is  used.  The  percentage  paid  for  each  year  is  called  the 
Rate  per  Cent.  The  interest  and  principal  added  together  form  the 
Amodnt. 

25.  What  is  the  interest  of  $137.84  for  1  year  at  6  per 
cent  ? 

26.  What  is  the  interest  of  $487.31  for  6  months  at  6  per 
cent  per  year? 

Sugyestion.  —  Since  the  rate  for  1  year  is  6  per  cent,  the  rate  for  6 


178  FKA(!TION3. 

months  must  be  j  of  6  per  cent,  which  is  3  per  cent.    The  interest  for  6 
months  will,  therefore,  be  3  per  cent  of  the  principal. 

27.  Mr.  Adams  borrowed  of  Mr.  Wales  $718.63,  for  Which 
he  agreed  to  pay  interest  at  6  per  cent.  At  the  end  of  6 
months  he  paid  the  principal  and  interest  How  much  did 
he  pay  ? 

28.  What  is  the  interest  of  $47.83  for  4  months  at  6  per 
cent  ? 

134.    Multiplication  and  Division  of  the  Numerator. 

The  foregoing  illustrations  have  shown  that  multiplying 
the  numerator  of  a  fraction  multiplies  the  fraction,  and  that 
dividing  the  numerator  divides  the  fraction.  The  same  thing 
may  be  demonstrated  more  rigidly,  by  considering  the  nature 
of  fractions,  thus  :  — 

Since  the  numerator  of  a  fraction  shows  how  many  parts 
are  considered,  multiplying  or  dividing  the  numerator  multi- 
plies or  divides  the  number  of  parts  considered,  without 
affecting  their  size,  and  hence  multiplies  or  divides  the  frac- 
tion. 

1.  Explain  the  effect  of  multiplying  the  numerator  of  the 
fraction  -^  by  3. 

Answer.  —  Multiplying  the  numerator  of  the  fraction  xV  by  3  gives 
W  for  a  result,  which  expresses  3  times  as  many  parts,  each  of  the 
same  size  as  before,  and  is,  therefore,  3  times  as  lai^e.  Hence,  the 
fraction  -^j  has  been  multiplied  by  3. 

See  Note  after  solution  of  example  8th. 

Explain  the  effect  of  multiplying  the  numerator  — 

2.  Of  T^  by  2.         4.   Of  f  by  4.  6.   Of  ^  by  9. 
8.   Oft^^byS.    I     5.   Of  §f  by  11.       7.   Of  ^f  by  8. 
8.    Explain  the  effect  of  dividing  the  numerator  of  the 

fraction  |  by  4. 

Solution.  —  Dividing  the  numerator  of  the  fraction  f  by  4  gives  ^ 
for  a  result,  which  expresses  ^  as  many  parts,  each  of  the  same  size  as 
before,  and  is,  therefore,  |-  as  large.  Hence,  the  fraction  f^  has  beea 
divided  by  4. 


FRACTIONS.  179 

Note.  —  The  work  explained  above  is  really  equiva^'^nf;  to  3  times 
^  = -{t^  (j^^st  as  3  times  5  apples  are  15  apples;)  to  :j-  of  f-  =  §» 
(just  as  4  of  8  apples  =  2  apples.)  The  above  forms  are,  however, 
necessary,  and  should  therefore  be  mastered. 

Explain  the  effect  of  dividing  the  numerator  — 


9.  Of  If  by  8. 
10.  Of  if  bj  2. 


11.  Of  ^f  by  9. 

12.  Of  if  by  3. 


13.  Of  if  by  6. 

14.  Of  e  by  12. 


130.    Multiplication  of  the  Denominator. 

{^a.)  Since  the  denominator  of  a  fraction  shows  into  how 
many  parts  the  unit  is  divided,  or  how  many  parts  equal  the 
unit,  multiplying  the  denominator  must  (131,  h.  and  c.) 
divide  each  part,  and  therefore  must  divide  the  fraction. 

1.  Explain  the  effect  of  multiplying  the  denominator  of 
the  fraction  ^  by  2. 

Answer.  —  Multiplying  the  denominator  of  the  fraction  £  by  2  gives 
f  for  a  result,  which  expresses  the  same  number  of  parts,  each  ^  as 
large  as  before.  Therefore,  f  =  2"  of  f ,  or  multiplying  the  denomina- 
tor of  f  by  2,  has  divided  the  fraction  by  2. 

Explain  the  effect  of  multiplying  the  denominator  — 

2.  Of  t  by  3.  5.   Of  i  by  7.  8.   Of  4  by  7. 

3.  Of  f  by  4.  6.    Of  f  by  2.  9.   Of  i^  by  10. 

4.  Of  f  by  4.  7.    Of  f  by  3.  10.    Of  ^  by  6. 
(6.)    Hence,  multiplying   the  denominator   of  a  fraction 

divides  the  fraction,  hy  dividing  the  size  of  each  part,  with" 
out  affecting  the  number  of  parts  considered. 

13G*    Division  of  the  Denominator. 

Since  the  denominator  of  a  fraction  shows  into  how  many 
Darts  the  unit  is  divided,  or  how  many  such  parts  are  equal 
to  the  unit,  dividing  the  denominator  must  (121,  h.  and  c.) 
multiply  each  part,  and  therefore  multiply  the  fraction. 

1.  Explain  the  effect  of  dividing  the  denominator  of  the 
fraction  }^  by  4. 

Ansictr.  —  Dividing  the  denominator  of  the  fraction  \^  by  4  gives 


180  FRACTIONS. 

-V"  for  a  result,  which  expresses  the  same  number  of  parts,  each  4  times 
as  large  as  before.  Therefore,  -V"  =  4  times  ^g-,  or,  dividing  the 
denominator  of  ^^  by  4  has  multiplied  the  fraction  by  4. 

Explain  the  effect  of  dividing  the  denominator  — 


2.  Off  by  3. 

5.  Of /^  by  5. 

3.  Of  I  by  4. 

6.  Of  ^1  by  12. 

4.  Of  ^  by  2. 

7.  Of  f  1  by  36. 

8. 

Of 

^f  by  6. 

9. 

Of 

^u  by  5. 

10. 

Of 

/^  by  3. 

Hence,  dividing  the  denominator  of  a  fraction  multiplies 
the  fraction^  hy  multiplying  the  size  of  each  part  without 
affecting  the  number  of  parts  expressed. 

137.    Recapitulation  and  Inferences. 

(a.)  Multiplying  the  numerator  multiplies  the  fraction^  hy  multiplying  the 
number  of  parts  considered  without  affecting  their  size. 

(h.)  Dividing  the  numerator  divides  the  fraction,  by  dividing  the  number 
of  parts  considered  without  affecting  their  size. 

(c.)  Multiplying  the  denominator  divides  the  fraction,  by  dividing  each 
part  without  affecting  the  number  of  parts  considered. 

(d.)  Dividing  the  denominator  multiplies  the  fraction,  by  multiplying  each 
part  without  affecting  the  number  of  parts  considered. 

{e.)  Hence,  1.  A  fraction  may  be  multiplied  either  by  midtiplying  the 
numerator  or  hy  dimding  the  denominator. 

2.  A  fraction  may  be  divided  either  hy  dividing  the  numerator  or  by 
multiplying  the  denominator. 

3.  Multiplying  both  numerator  and  denominator  of  a  fraction  by  any 
lumber  both  multiplies  and  divides  the  fraction  hy  that  number,  and,  there- 
fore, does  not  alter  its  value. 

4.  Dividing  both  numerator  and  denominator  of  a  fraction  hy  the  sa7ne 
number  both  divides  and  multiplies  the  fraction  by  that  number,  and,  there- 
fore, docs  not  alter  its  value. 


138.    Multiplication  and  Division  of  both  Kumeraior  and 
Denominator  by  the  same  Number. 

1.    Explain  the  effect  of  multiplying  both  numerator  and 
denominator  of  the  fraction  \  by  6. 

Answer.  —  Multiplying  both  numerator  and  denominator  of  the  frao* 


FRACTIONS.  181 

tion  f-  by  6  gives  f  f  for  a  result,  which  expresses  6  times  as  many 
parts,  each  ^  as  large  as  before.  Therefore,  the  value  of  the  fraction  is 
unaltered,  and  j  =  f  f . 

Explain  the  effect  of  multiplying  both  numerator  and  de- 
nominator of — 


2.  The  fraction  .§  by  2. 

3.  The  fraction  ^  by  9. 

4.  The  fraction  ^^  by  10. 

5.  The  fraction  §|  by  7. 


6.  The  fraction  ^\  by  3. 

7.  The  fraction  /^  by  8. 

8.  The  fraction  ^f  by  9. 

9.  The  fraction  ^f  by  4. 


10.  Explain  the  effect  of  dividing  both  numerator  and  de- 
nominator of  the  fraction  -f^  ^7  ^* 

Answer.  —  Dividing  both  numerator  and  denominator  of  the  fraction 
r2"  by  3  gives  f  for  a  result,  which  expresses  ^  as  many  parts,  each 
part  3  times  as  large  as  before.  Therefore,  the  value  of  the  fraction  is 
unaltered,  and  -^^  =  f. 

Explain  the  effect  of  dividing  both  numerator  and  denomi- 
nator of  — 


11.  The  fraction  Jf  by  8. 

12.  The  fraction  |i  by  9. 

13.  The  fraction  ^\  by  8. 

14.  The  fraction  ^f  by  18. 


15.  The  fraction  y^^  by  7. 

1 6.  The  fraction  -^^^^  by  41. 

17.  The  fraction  f^  by  5. 

18.  The  fraction  ^\%  by  39. 


130.    Lowest  Terms. 

(a.)  The  numerator  and  denominator  are  called  terms  of 
the  fraction. 

(h.)  A  fraction  is  said  to  be  reduced  to  its  lowest  terms 
when  its  numerator  and  denominator  are  the  smallest  entire 
numbers  which  will  express  its  value. 

(c.)    From  the  preceding  explanations,  it  follows  that,  — 

1.  A  fraction  may  be  reduced  to  lower  terms  by  dividing 
both  numerator  and  denominator  by  any  number  which  will 
divide  both  without  a  remainder. 

2.  A  fraction  may  be  reduced  to  its  lowest  terms  by  divid- 
ing both  numerator  and  denominator  by  any  number  which 

16 


182  FRACTIONS. 

will  divide  both  without  a  remainder ;  then  dividing  this 
result  in  the  same  way,  and  so  on,  continuing  the  process  till  a 
fraction  is  obtained,  the  terms  of  which  are  prime  to  each 
other ;  or  by  dividing  both  numerator  and  denominator  by 
their  greatest  common  divisor. 

3.  A  fraction  will  always  be  reduced  to  its  lowest  terms 
when  there  is  no  number  greater  than  1  which  will  divide 
both  its  numerator  and  denominator  without  a  remainder. 

1.  Reduce  y^j  to  its  lowest  terms. 

Solution.  —  Dividing  both  numerator  and  denominator  by  4,  their 
greatest  common  divisor,  gives  f ,  which  expresses  ^  as  many  parts, 
each  part  4  times  as  large  as  before.     Hence,  -^^  =  f-. 

2.  Reduce  ffl^l  to  its  lowest  terms. 

Solution.  —  Observing  (104,  11.)  that  both  numerator  and  denomi- 
nator are  divisible  by  4,  we  first  divide  by  4,  which  gives  t^^Wt?  or  i 
as  many  parts,  each  4  times  as  large  as  before. 

Observing  (104,  IV.)  that  both  numerator  and  denominator  of  the 
last  fraction  are  divisible  by  9,  we  divide  by  9,  which  gives  "i^fy,  or  g- 
as  many  parts,  each  9  times  as  large  as  before. 

Observing  (104,  V.)  that  both  numerator  and  denominator  of  the 
last  fraction  are  divisible  by  11,  we  divide  by  11,  which  gives  -^sV?  or 
l^X  i^s  many  parts,  each  11  times  as  large  as  before.  As  the  numerator 
and  denominator  of  the  last  fraction  are  prime  to  each  other,  the  reduc- 
tion can  be  carried  no  farther,  and  f  §f  ^f  reduced  to  its  lowest  terms 
equals  tVt- 

Second  Solution.  —  The  greatest  common  divisor  of  38412  and  50292 
found  by  one  of  the  methods  of  Section  IX.)  is  396  ;  and  dividing  both 
numerator^ and  denominator  by  it,  gives  i^^V^  or  r^i^  as  many  parts, 
each  part  396  times  as  large  as  before.     Hence,  f  a|^|  =  -^j- 

NoTB.  —  The  mechanical  procesa  by  the  first  solution  is  merely  to 
divide  both  terms,  first  by  4,  then  by  9,  then  by  U  ;  while  by  the  last  it 
is  to  find  the  G.  C.  D.  of  both  terms,  and  divide  them  by  it.  The  first 
method  will  usually  be  the  most  convenient,  when  the  divisors  can  be 
readily  perceived. 

(d.)  The  pupil  should  be  careful  not  to  decide  that  any 
fraction  is  incapable  of  reduction  till  he  has  carefully  tested 
it  by  some  of  the  processes  of  Section  IX. 


FRACTIONa. 


185 


Reduce  each  of  the  following  fractions  to  its  lowest  terms. 


3.  U- 

4.  if. 

5.  iM. 

6.  ih 


7.     M.* 


8. 

H- 

9. 

il 

10. 

iU- 

11-  Iff- 

12.  IJii- 

13.  Iff. 
14-  i^s^A- 


14L0.     Cancellation, 

(a.)  When,  as  is  sometimes  the  case,  the  factors  of  the 
flumerator  and  denominator  are  given,  labor  may  be  saved  by 
reducing  the  fraction  to  its  lowest  terms  before  multiplying 
the  factors  together. 

(b.)  In  writing  the  work,  it  is  well  to  draw  a  line  through 
the  factors  which  have  been  divided,  and  to  place  the  quotients 
above  those  in  the  numerator,  and  beneath  those  in  the  de- 
nominator. 

(<?.)  The  numbers  by  which  we  divide  are  said  to  be  can- 
celled, and  the  process  is  called  cancellation  ;  but  it  is  identi- 
cal in  principle  with  other  cases  of  reducing  fractions  to  their 
lowest  terms. 


1.   Reduce 


8  X  15 

9  X  16 


to  its  lowest  terms. 


Solution.  —  Cancelling  8  from  the  factors  8  of  the  numerator  and  16 
of  the  denominator,  (i.  e.,  dividing  each  by  8,)  gives  1  in  place  of  8, 
and  2  in  place  of  16,  and  makes  the  fraction  express  ^  as  many  parts, 
each  8  times  as  large  as  before.f 

Cancelling  3  from  the  factors  15  of  the  numerator  and  9  of  the  de- 
nominator, gives  5  in  place  of  15,  and  3  in  place  of  9,  and  makes  the 
fraction  express  ^  as  many  parts,  each  3  times  as  large  as  before.J  As 
no  further  division  can  be  made,  the  remaining  factors  are  to  be  multi- 
plied together,  which  gives  |  for  a  result. 


*  Solution.  64  and  81  are  prime  to  each  other,  and  hence  f  *  cannot  be 
reduced  to  lower  terms, 

t  For  multiplying  by  |-  of  a  number  gives  ^  as  large  a  product  as  mul- 
tiplying by  the  number  would  give. 

X  For  multiplying  by  ^  of  a  number  gives  -g-  as  large  a  product  as  mul- 
tiplying by  the  number  would  give. 


184  FRACTIONS. 

The  work  would  be  written  thus  :  — 
1  5 

0  X  ^^         5 


3  2 

„     ^   ,        12  X    7    X  25  X  36  X  11      .    , 

2.  Reduce  35  ^  12  X    4  X  H  X  21  *^  ^^  lowest  terms. 

Solution.  —  Cancelling  the  factor  12  from  numerator  and  denomina- 
tor, gives  1  in  place  of  each.  .Cancelling  7  from  the  numerator  and 
from  the  35  in  the  denominator,  gives  1  in  place  of  the  former,  and  5  in 
place  of  the  latter.  Cancelling  5  from  the  denominator  and  from  the 
25  in  the  numerator,  gives  1  in  place  of  the  former,  and  5  in  place  of 
the  latter.  Cancelling  4  from  the  denominator  and  from  the  36  in  the 
numerator,  gives  1  in  place  of  the  former,  and  9  in  place  of  the  latter. 
Cancelling  11  from  the  numerator  and  denominator,  gives  I  in  place  of 
each.  Cancelling  3  from  the  9  in  the  numerator  and  from  the  21  in  the 
denominator,  gives  3  in  place  of  the  former,  and  7  in  place  of  the  latter. 
As  no  further  reduction  can  be  made,  we  multiply  the  remaining  factors 
together,  which  gives  -7-^-  =  2^. 

The  written  work  would  be  thus  :  — 

3 
115  0  1 

X^  X  :t  X  ^$  X  M  X  :it  _  15  _   1 
^^  X  ^^  X  ^  X  i^^  X  ^^      7-7 

^1117 

1 

Or,  by  omitting  to  write  the  factors  which  are  equal  to  1,  as  we  may 
do  without  ambiguity,  we  have  the  following  more  convenient  form. 

3 
5  0 

x^  X  ;t  X  ^$  X  U  X  :(t  _i5_i 
fi$  X  t^  X  i  X  Xi  X  M~  1  ^   1 

$  7 

o    r»  J  5  X  7  X  11  X  12  X  15  X  18  X  2       ^ 

3.  Reduce  T-rr-^-zz—ci 


4X5X3x11X7X36X5X6 

its  lowest  terms. 


i 


FRACTIONS.  '  185 


Aru 

$X^ 
X  $  X 

Reduce 

X  n 

X  i^ 

X  x$ 

X  U 

X 

^ 

1 

4: 

4. 

^    X  XX  X    "^    X  M  X 
48  X  30  X  49  X  64  X  27 

to 

X 

0-2 
2 

!  lowest 

21  X  32  X  36  X  42 
terms. 

Answer.  —  3 

4.  10  ^  ^  0 

i$  X  i0  X  4^  X  U  X  ^"^  _  120  _ 

^^  X  $^  X  ;gl0  X  ^^  1 

;r  ;^       0 

Reduce  each  of  the  following  to  its  lowest  terms. 
7   X16X18X5X   9 


120 


10.  — 
11 


20  X  14  X  9  X  9  X  11 
8  X  36  X  28  X  48  ^ 
24  X  72  X  13  X  15 

6X7  X  8  X  9  X  10  X  11  X  12 
7  X  8  X  9  X  10  X  11  X  12  X  IB 

96  X  65  X  35 
25^"90r48   " 

14  X  5  X  9  X  30  X  16  X  17 
8  X  17  X  14  X  27  X  11  X  3 

72  X  49  X  81  X  33 


77  X  84  X  6  X  27  X  99 
144  X  103  X  625  X  121 


12. 
13. 


375  X  64  X  99  X  847 

315  X  143  X  64  X  221 
119  X  169  X  209  X  125 

1331  X  343  X  6859 
1463  X  2527  X  847 
16* 


166  FRACTIONS. 

U  4959  X  3487  X  2491 

6061  X    53    X  2853  X  47 


14-1«    Compound  Fractions. 

(a.)  A  COMPOUND  FRACTION  IS  a  fraction  of  a  fraction 
as  f  of  ^,  f  of  I  of  ■^%. 

(b.)  A  compound  fraction  is  equivalent  to  a  fraction  mul- 
tiplied by  a  fraction. 

Thus :  f  of  f  =  J  times  y ;  f  of  f  of  y®^  =  f  times  |  times 
rV     (See  132.) 

(c.)  In  a  compound  fraction,  the  value  expressed  by  one  fraction  is 
made  the  unit  of  another,  f  of  ^  means  f^  of  the  quantity  ^,  or  3  such 
parts  as  would  be  obtained  by  dividing  -f  into  4  equal  parts,  -j  of  f  of 
Y^xr  means  §  of  the  quantity  ^  of  x^u,  and  ^  of  -j^jy  means  |-  of  the 
quantity  j^j. 

(d.)  In  reducing  compound  fractions  to  simple  ones,  it  is  important 
to  notice  which  fraction  is  made  the  unit  of  the  other,  as  that  is  the  one 
on  which  the  operation  is  to  be  performed. 

1.   Whatisf  of  tHI? 

Solution,  f  of  f  11^  =  8  times  ^  of  ff |^ ;  i  of  f  JH,  found  by 
dividing  5747  by  9,  is  ^|f  f  ^,  and  8  times  this  result,  found  by  multiply- 
ing 638f-  by  8,  is -.    Hence  the  following  forms  of  written  work. 

First  Form.  Second  Form. 
a  =  5747  9  )  5747    • 

^  of  a  =  b  =  essf  '    — 

8  times  b  =  5108f  =  f  of  a. 

Hence,  f  of  tJH  =  tHI* 


5108| 

Hence,|ofHH  =  liM* 

2.  What  is  yV  of  f  f  f  ?  5.   What  is  ^  of  |Jf  ? 

3.  What  is  I  of  ^^V^  ?  6.   What  is  ^  J  of  ||f  ^f  ? 

4.  What  is  -i^  of  l^f  ^  ?  7.   What  is  f  |  of  |f  ^f  ? 

8.   What  is  ^  of  437^  ? 

Solution.      -^  of  437f  =  54,  with  a  remainder  of  5^,  which,  reduced 
tc  sevenths.,  =  V^;  ^  of  \<i  =  f.     Hence,  i  of  437^  =  54^. 


FRACTIONS.  187 


Note. —  Compare  the  above  with  "  What  is  J  of  437  weeks  and  5 
days  ?  " 


9.   What  is  ^  of  8539^  ? 

10.  What  is  f  of  827|  ? 

11.  What  is  ^  of  5827J  ? 


12.  What  is  I  of  2793  J- ? 

13.  What  is  I  of  1397-J-j-  ? 

14.  What  is  I  of  4355^^1? 


Second  Method  of  Reduction, 

(e.)  When  the  numerator  of  the  unit  of  tlie  compountl 
fraction  is  not  a  multiple  of  the  denominator  of  the  other 
fraction,  the  preceding  solution  will  give  a  fraction  in  the 
numerator  of  the  result.  In  all  such  cases  the  following  solu- 
tion should  be  adopted.  Indeed,  applying,  as  it  does,  the 
principles  of  cancellation,  it  really  includes  the  preceding. 

15.    Reduce  |  of  |^  to  a  simple  fraction. 

First  Solution. —  Since  f  of  any  fraction  must  equal  3  times  as  many 
parts,  each  ^  as  large  as  before,  f  of  |-  may  be  found  by  multiplj-ing 
the  denominator  of  |-  by  8,  and  the  numerator  by  3.  This  gives  the 
following  written  work. 

^  o{-   =  ^  ^  ^         1 


8       9         0X0         6 
3        2 

Second  Solution,  i^  of  -g-  may  be  expressed  by  making  8  a  factor  of 
the  denominator,  and  f  of  |-  must  be  3  times  this  result,  which  may  be 
expressed  by  making  3  a  factor  of  the  numerator.  This  gives  the  same 
written  work  as  before. 

Note.  —  In  writing  the  work  of  such  examples  as  the  above,  the 
fraction  on  which  the  operation  is  to  be  performed  should  always  be 
written  first. 

16.    Reduce  |  of  ^|  of  3^  to  a  simple  fraction.    » 

First  Solution.  3^,  or  -f-f-,  being  the  number  on  which  the  opera- 
tion is  to  be  performed,  should  be  written  first.  ^|  of  this  must  equal 
15  times  as  many  parts,  each  -J^  as  large,  and  may  be  expressed  by 
making  15a  factor  of  the  numerator,  and  28  a  factor  of  the  denominator, 
I  of  this  must  equal  8  times  as  many  parts,  each  ^  as  large,  and  hence 
may  be  expressed  by  making  8  a  factor  of  the  nvunerator,  and  9  a  factor 
of  the  denominator.     This  would  give  the  following  work. 


188  FRACTIONS. 

5  5^ 

9  °    28  ^      11         11  X  ^^  X  0         33  33 

4         3 

Second  Solution.  S-^y,  or  ^^,  is  the  number  on  which  the  operation 
is  to  be  performed,  and  should  therefore  be  written  first.  2-*^  of  j\ 
may  be  expressed  by  making  28  a  factor  of  the  denominator,  and  ^| 
of  j^  must  be  15  times  this  result,  which  may  be  expressed  by  making 
15  a  factor  of  the  numerator.  ^  of  this  result  may  be  expressed  by 
making  9  a  factor  of  the  denominator,  and  f-,  or  8  times  the  last  result, 
by  making  8  a  factor  of  the  numerator.  This  would  give  the  same 
written  work  as  before. 

Note.  —  It  will  be  seen  that  both  of  the  above  solutions  give  the 
same  numerical  process,  viz.,  to  make  all  the  numerators  of  the  com- 
pound fraction  factors  of  the  new  numerator,  and  all  the  denominators 
factors  of  the  new  denominator,  and  then  cancel  and  reduce. 

Reduce  each  of  the  following  fractions  to  simple  ones. 


17.    iofff 

22.     iV  of  A  of  13^. 

18.     ^fofH- 

23.     fof^ofS}. 

19.     fof^f. 

24.    fofy^yofi^. 

20.    fof^^^ofl. 

25.     i|ofifof4ff 

21.     foffofH- 

26.     ^  of  i  of  f  of 

iof^fofff 

27.    ^^of^o£^\ 

of  j\  of  9^. 

28.    1  of  T^^  of  II 

of  H  of  If 

29.     1  of  ^  of  fl  of  If  of  13i. 

30.     f  of  1  of  f  of 

Sof 

^  of  H  of  if  of  ft  of  62? 

14:S.    Multiplication  of  Fractions, 
1.   What  is  the  product  of  |  X  f  ? 

Solution.  I  multiplied  by  t  =  I  of  |,  which,  found  by  multiplying 
the  numerator  of  i  by  3  and  the  denominator  by  8,  gives  the  following 
written  work. 

_4         3  _  iX  ^  ^  1^ 
9        8         0  X  $         6 
3        % 


FRACTIONS. 


189 


(a.)    Bj  reading  the  sign  X  as  timeSf  we  have  — 

I"  times  f  =  f  of  f ,  which,  found  by  multiplying  the  numerator  of 
f  by  4  and  the  denominator  by  9,  gives  the  following  written  work. 


9^8- 

^  X  4        1 

^  X  0         6 
2       3 

What  is  the  product  of — 

2.  fXM? 

3.  AVXtl? 

4.  ^XUX^\? 

5.  f  Xf  X#? 

6.  4^  X  4f  X  8i  ? 

7.  3i  X  6^  X  T^^  ? 

8.  13f  X  23^7^  X 

9.  5^X1  of  St^t  ? 
10.    t  X  f  X  iJ  X 

J  7  3    ? 
'TITS-  • 


§S? 


(5.)  The  following  form  of  explanation  may  be  adopted 
when  a  very  thorough  analysis  is  required. 

11.   What  is  the  product  of  AJ  x  f  ^  ? 

Solution.  —  First  write  ^f^  as  the  number  on  which  the  operation 
is  to  be  performed ;  we  then  have  f  f-  multiplied  by  1  equals  f  f ,  and 
multiplied  by  ^V  wiU  equal  g^  of  this  result,  which  may  be  expressed 
by  making  98  a  factor  of  the  denominator."  If  multiplying  by  ^V  gives 
this  result,  multiplying  by  f  |-  must  give  81  times  this  result,  which  may 
be  expressed  by  making  81  a  factor  of  the  numerator.    Hence,  — 

3 

49         SI   _  4^  X  $i         S 
54 


81   _ 
^   98  - 


$4  X  0^ 

2  2 


Note.  —  It  will  be  seen  that  all  the  forms  of  solution  give  similar 
forms  of  written  work,  the  numerators  of  the  several  fractions  being  in 
all  cases  factors  of  the  numerator  of  the  product,  and  the  denominators 
factors  of  the  denominator. 


What  is  the  product  of — 
12.     fX-Z^-XH? 

13.  uxnxm^ 

14.     HXAXil? 


15-     1%%  X  tttWXtVtt? 

16.  .83  X  .007  X  .49?* 

17.  8.7  X  .43  X  .006?* 


*  For  models  of  written  work,  see  171st  page,  solution  to  8th  example, 
and  172d  page,  Note,' 


I9t)  FRACTIONS. 

18.  4f  X  2|  X  9i  ?         I     20.    .824  X  4.8  ? 

19.  43.79  X  25.7  ?  I     21.     6750  X  .6750  ? 

22.  5.06  X  300  X  .2  X  .0004  ? 

23.  .4  X  .3  X  .2  X  .6  X  .2  ? 

24.  4.974  X  1.0007  ? 

143.    Beditction  of  a  Vulgar  Fraction  to  a  Decimal  Form 

1.  Reduce  |  to  a  decimal  fraction. 

Solution.  ^,  or  |-  of  1,  =  i  of  7  =  |-  of  70  tenths  =  8  tenths,  with 
a  remainder  of  6  tenths.  But  6  tenths  =  60  hundredths,  and  ^  of  60 
hundredths  =  7  hundredths,  with  a  remainder  of  4  hundredths.  But  4 
hundredths  =  40  thousandths,  and  ^  of  40  thousandths  =  5  thou- 
sandths.   Hence,  |^  =  .8  -f-  .07  +  .005  =  .875 

The  work  may  be  written  thus  :  — 

8  )  7-000  Proof.     .875  ==  AV(T  =  f 

.875 

2.  Reduce  ||  to  a  decimal  fraction. 

Solution,    f f,  or  f  f  of  1,  =  sV  of  25  =  sV  of  250  tenths  =  8 

tenths,  with  a  remainder  of  26  tenths.    But  26  tenths  =  260  hundredths, 

and  2V  of  260  hundredths  =  ;  &c.    Hence  the  following  written  work. 

28  )  25.00  .8928572V  =  ^ns.,  or  dropping  the  ^, 

2.60  or  y  of  a  millionth,  we  shall  have  the 

•  approximate  value  of  §t  =  .892857 

.080  ^ 


.0240 

.00160 

.000200 


.000004 
Proof.      .892857^    =   892857f   times   ttt^^tt  =  ^^^2^^  of 

Note. —  Compare  the  above  soHitions  with  8T,  (c.) 

(5.)    The  fractions  may  also  be  reduced  by  the  following 
solutions :  — 

1^,  or  ^  units,  =  10  times  as  many  tenths  =  — g tenths  =  10 

times  as  many  hundredths  as  tenths  =  <  X  10  X  10  hm^^j-edths  =*  10 


FRACTIONS.  191 

times  as  many  thousandths  as  hundredths  =  .  J^ — 1^ —  thou- 
sandths. 


thousandths  =  .875 


By  cancelling  the  last,  we  have,  — 
5  5  5 

7  _  7  X  3^0  X  3:0  X  3^0 

i 

Note.  —  The  ahove  is  equivalent  to  the  following :  — 
5  5  5 

7         1  X  t0  X  t0  X  :i0          875 


8        $  X  10  X    10  X  10  ""  1000 
4 


=  .875 


25         25       .  .  ,  25  X   10         , 

So,  — ,  or  —  units,  =  10  times  as  many  tenths  =  — — —  tenths, 

28  28  28 

,       ^     ,  L  .      ,.  25  X   10  X   10  ^ 

s=  10  times  as  many  hundredths  as  tenths  = hun- 

*'  28 

dredths,  &c. 

5         5 
Hence  ^  -  ^^  X  3:0  X  3:0  X   10  X   10  X   10  X  10 

'  28  —  ^0 

7 
millionths  =  —^ millionths  c=  .892857^ 

Note.  —  The  above  is  equivalent  to 

5    5 
25^  _  25  X  3^0  X  3^0  X  10  X  10  X  10  X  10  _ 
^  ~~  ^$  X  10  X  10  X  10  X  10  X  10  X  10  ~ 

u 

7 

'^^-'1^0^  .000001  =  .S92S67^ 
7000000  7  ^ 

(c.)   It  is  obvious  that  f  |  cannot  be  reduced  to  an  exactly 
equivalent  decimal  form,  for  introducing  into  the  numerator 


Id2  FRACTIONS. 

any  number  of  factors  each  equal  to  10  will  not  enable  us  to 
cancel  the  factor  7  from  the  denominator.  As  the  same  prin- 
ciples apply  to  any  other  fraction,  it  follows  that,  — 

{d.)  No  vulgar  fraction  can  he  reduced  to  an  exactly  equiv- 
alent decimal  form,  if  when  reduced  to  its  lowest  terms,  its 
denominator  contains  other  factors  than  such  as  are  found  in 
10,  i.  e.,  other  than  2's  or  5*s  ;  and  conversely,  that,  — 

(e.)  Every  vulgar  fraction  which,  when  reduced  to  its  low- 
est terms,  contains  in  its  denominator  only  such  factors  as  are 
found  in  10,  can  he  reduced  to  an  exactly  equivalent  decimal 
form,  and  will  contain  as  many  decimal  places  as  there  are 
2*5  or  5'*  to  he  cancelled  from  the  denominator, 

15         15 

Illustrations.     —  =  — »  and  hence  can  be  reduced  to  an  exactly 

equivalent  decimal.  Moreover,  it  will  contain  four  decimal  places ;  for 
10  must  be  introduced  4  times  as  a  factor  into  the  numerator,  to  cancel 
2<  from  the  denominator. 

13  13 

■Again.    -— —  =  r-rr-To'  and  hence  can  be  reduced  to  an  exactly 
250         2X5^  " 

equivalent  decimal.  Moreover,  it  will  contain  3  decimal  places,  for  10 
must  be  introduced  3  times  as  a  factor  in  the  numerator,  to  cancel  the 
factors  of  the  denominator. 

27  27 

Again.    —-  =  ■>  and  hence  cannot  be  reduced  to  an  exactly 

OJi  2r  X    lo 

equivalent  decimal  form. 

(/.)  Vulgar  fractions  which  cannot  be  exactly  reduced 
give  rise  to  repeating  or  circulating  decimals. 

Reduce  each  of  the  following  to  a  decimal  form,  carrying 
the  division,  when  only  approximate  values  can  be  obtained, 
to  six  decimal  places. 


3.  iii- 

7.    A- 

11.    §f. 

4-   4i. 

8.    if. 

12.    H- 

5.    Ji. 

9-    tft. 

13.    if 

6.    \\.              \ 

10.    f 

14.    iJ. 

14:4:.    Fractional  Parts  of  Denominate  Numbers, 
(a.)  What  is  the  value  of  f  of  a  mile  in  whole  numbers  of 


lower  denominations,  i.  e.,  in  furlongs,  reds,  yards,  &c. 


? 


FKACTIONS.  195 

First  Solution. —  This  example  may  be  solved  by  the  common  pro- 
cess of  compound  division,  thus  :  |-  of  a  mile  =  ^  of  8  miles  =  0 
miles,  with  8  miles  remaining.  But  8  miles  =  64  furlongs,  and  ^  of  64 
furlongs  =  7  furlongs,  with,  &c. 

Second  Solution.  —  Since  1  m.  =  8  fur.,  |-  of  a  mile  must  equal  f- 

of  8  fur.,  which  is  7^  fur.     Since  1  fur.  =  40  rods,  ^  of  a  fur.  must 

equal  ^  of  40  rods,  which  is  4^  rods.     Since  1  rod  =  5^,  or  -^^-  yds.,  |- 

of  a  rod  must  equal  |-  of  ^2^-  yds.,  which  is  2^  yds.     Since  1  yd.  = 

3  ft,  f  of  a  yd.  must  equal  J-  of  3  ft,  which  is  l|-  ft.     Since  1  ft  = 

12  in.,  ^  of  a  ft.  must  equal  ^  of  12  in.,  which  is  4  in.     Therefore,  § 

of  a  mile  =  7  fur.,  4  rd.,  2  yd.,  1  ft.,  4  in.     The  work  may  be  written 

thus : — 

m.  fur,  fur.  fur.  rd.  rd. 

8        8  X   8        64        .1                 1         1   X   40         40  4 

•^  =  — :: —  =  -:r,  or  7r  -  =    — =  -tt,  or  4- 


or  1- 


8 

X 

8 

9 

yd. 

4 

X 

11 

9 

X 

in. 

2 

1 

X 

12 

64     1 

1 
9 

yd. 

yd. 

22    4 

4 
9 

rd.        yd.  yd.  yd.  ft. 

4         4X11         22  4  4         4X3         4 


99X2  9'9  9  9  3'         3 

=  4  in.     Hence,  f  m.  =  7  fur.  4  rd.  2  yd.  1  ft.  4  in. 


ft. 
1 
3  3 

Third  Solution.  —  Since  there  are  8  furlongs  for  every  mile,  there 
must  be  8  times  as  large  a  part  of  a  furlong  as  of  a  mile,  or,  in  this  in- 
stance, 8  times  g-  of  a  furlong,  which  is  7^  furlongs.  Since  there  are  40 
rods  for  every  furlong,  there  must  be  40  times  as  large  a  part  of  a  rod 
as  of  a  furlong,  or,  in  this  instance,  40  times  g-  of  a  rod,  which  is  4^ 
rods.  Since  for  every  rod  there  are  5^,  or  -^  yards,  there  must  be  -^ 
as  large  a  part  of  a  yard  as  of  a  rod,  or,  in  this  instance,  ^^  of  ^  yard, 
which  is,  &c.     The  written  work  would  be  the  same  as  before. 

(6.)  The  only  difference  between  the  processes  of  this  and  the  pre- 
vious article  is  the  difference  between  decimal  and  denominate  numbers. 
In  the  former,  a  unit  of  any  denomination  equals  10  of  the  next  lower, 
while  in  the  latter  the  number  of  units  of  a  lower  denomination  to 
which  any  unit  is  equal,  varies  with  the  denomination  of  the  unit  con- 
Bidered. 

What  is  the  value  of  each  of  the  following  in  whole  num- 
bers of  lower  denominations  ? 


1. 

f  of  a  £. 

6. 

y\  of  an  acre. 

2. 

^^  of  a  gal. 

7. 

^1  of  a  mile. 

3. 

^1^  of  a  bu. 

8. 

■^^  of  a  ton. 

4. 

f  f  of  a  week. 

9. 

A  of  a  lb.  T. 

5. 

ii  of  a  Cd.  ft. 

1; 

194  FRACTIONS. 

(c.)  Reduce  .7925  of  a  £  to  whole  numbers  of  lower  de- 
nominations. 

First  Solution.— Since  £1  =  20  s.,  .7925  of  a  £  must  equal  .7925 
of  20  s.  =  20  times  .7925  s.  =  15.8500  s.  Since  1  s.  =  12  d.,  .85  of  a 
8.  must  equal  .85  of  12  d.  =  12  times  .85  d.  =  10.20  d.  Since  1  d.  = 
4  qr.,  .2  of  a  penny  must  equal  .2  of  4  qr.  =  .8  qr.  Hence,  .7925  of  a 
iE  =  15  s.,  10  d.,  .8  qr. 


.7925  £. 
20 

WBITTEN  WORK. 

Or,  by  omitting  to  write  the  mul- 
tipliers, we  have,  — 

15.8500  s. 

.7925  £. 

12 

15.8500  s. 

10.20  d. 
4 

10.20  d. 

.8  qr. 

.80 

Hence,  .7925  of  a  £  =  15  s.  10  d.  8  qr. 

Second  Solution.  —  Since  there  are  20  s.  for  every  pound,  there  must 
be  20  times  as  large  a  part  of  a  shilling  as  of  a  pound,  or,  in  this  case, 
20  times  .7925  s.  =  15.85  s.  Since  there  are  12  d.  for  every  shilling, 
there  must  be  12  times  as  large  a  part  of  a  penny  as  of  a  shilling, 
or,  in  this  case,  12  times  .85  s.,  &c.  The  written  work  is  the  same  as  in 
the  last  solution. 

Note.  —  The  student  should  be  careful  to  multiply  only  the  fraction- 
1  part  of  each  number. 


"What  is  the  value  of — 

1.     .345  of  a  ton  ? 

5. 

.2376  of  a  gal.? 

2.     .6375  of  a  mile  ? 

6. 

.625  of  a  bu.  ? 

3.     .3426  of  a  lb.  ? 

7. 

.317  of  acu.  yd.? 

4.     .754  of  a  £.  ? 

8. 

.2569  of  a  cwt.  ? 

I 


9.  What  part  of  1  mile  is  7  fur.  4  rd.  2  yd.  1  ft.  4  in.  ? 

Solution.  —  Since  1  in.  =  yV  of  a  foot,  4  in.  must  equal  i^,  or  ^  of 
a  foot,  to  which  adding  the  1  foot  gives  1^  ft.  =  ^  of  a  foot.  Since 
1  ft.  =  ^  of  a  yd.,  4  of  a  foot  must  equal  ^  of  ^  of  a  yd.,  or  f  of  a  yd., 
to  which  adding  the  2  yards  gives  2|  yd.,  or  ^  yd.  Since  1  yd.  = 
^  of  a  rd.,  -^  of  a  yd.  must  equal  -V"  of  -n:  of  a  rd.,  or  |  of  a  rd.,  to 
which  adding  the  4  rd.  gives,  &c. 

When  the  work  is  written,  the  following  form  may  bo  adopted  :  — 


FRACTIONS, 
ft.                                               ft                    yd. 

'-  =  ^  =  j 

yd.                       rd. 

1        4        4    ,1        4 

■3=5  =  5»f3-=r 

rd.                        fur. 

2 
4  __  22  _>aL    .  2  _  4 
'9  —  9          9   ®  -H-       9 

4l  =  12='??ofl  = 
9         9          9       W- 

195 


fur. 


8 
1  _  64  _^  f  1  _  ^ 
9  "~  9  ~  "9  °  -Ss""  9 
Hence,  7  fur.  4  rd.  2  yd.  1  ft.  4  in.  =  f  of  a  mile. 
Proof.  —  Reduce  f  of  a  mile  to  fur.,  rd.,  &c. 

Second  Solution.  —  Since  there  is  yg"  of  »  foot  for  each  inch,  there 
must  be  ^  as  many  feet  as  inches,  or,  in  this  case,  xV  of  4  ft.  =  -j^,  or 
^  of  a  ft.  Since  there  is  -J  of  a  yard  for  each  foot,  there  must  be  ^  as 
many  yards  as  feet,  or,  in  this  case,  -g-  of  '-g-,  or  ^  yd.  =,  &c. 
The  written  work  is  the  same  as  in  the  last  solution. 
Note.  —  The  method  here  given  is  usually  preferable  to  that  of 
129,  solution  of  72d  example,  inasmuch  as  it  keeps  all  the  fractions 
reduced  to  their  lowest  terms,  and  enables  us  to  perform  very  many,  if 
not  most,  such  examples  without  writing  any  figures. 

What  part  — 

10.  Of  1  A.  is  2  E.  36  sq.  rd.  8  sq.  yd.  2  sq.  tt.  36  sq.  in.  ? 

11.  Of  1  gal.  is  3  qt.  1  pt.  11  gi.  ? 

12.  Of  1  Cd.  ft.  is  10  cu.  ft.  1382|  cu.  in.  ? 

13.  Of  1  T.  is  17  cwt.  3  qr.  2  lb.  12  02.  7^  dr. 

14.  Of  1  lb.  is  6  oz.  13  dwt.  8  gr.  ? 

15.  Of  1  £  is  9  s.  5  d.  li  qr.  ? 

16.  Of  1  circumference  is  155°  4'  36 jf"? 

17.  Ofl  lbis5g  6309  6|gr.?   ' 

18.  Of  1  w.  is  3  da.  10  h.  17  m.  84  sec.  ? 

(c?.)  Should  it  be  required  to  give  the  answers  to  such 
questions  as  the  above  in  a  decimal  form,  it  will  only  be 
necessary  to  reduce  the  vulgar  fractions  obtained  by  the  pre- 
ceding process  to  equivalent  decimals. 

(c.)  The  following  process  may  also  be  applied  : — 

19.  What  part  of  a  pound  is  13  s.  7  d.  2  qr.  ? 
SbZuiton.— Since  for  every  farthing  there  is  J  of  a  penny,  there  must 


19C  FRACTIONS. 

be  ^  as  many  pence  as  farthings,  or,  in  this  case,  ^  of  2  d.  =  .5  of  a 
penny,  to  which  adding  the  7  d.  gives  7.5  d.  Since  for  every  penny 
there  is  iV  o^  ^  shilling,  there  must  be  yg-  as  many  shillings  as  pence, 
or,  in  this  case,  tV  of  7.5  s.  =,  &c. 

The  most  convenient  form  of  writing  the  work  is  to  arrange  the 
numbers  expressing  the  various  denominations  in  a  vertical  column, 
thus:  — 

4  )    2.0  qr. 

12)    7.500  d.  =  7  d.  2  qr. 

20  )  13.6250  s.  =  13  s.  7  d.  2  qr. 

.68125  £  =  13  s.  7  d.  2  qr.  =  Am. 

In  like  manner  perform  the  following  : — 

20.  What  part  of  1  bu.  is  5  pk.  3  qt.  1  pt.  ? 

21.  What  part  of  1  lb.  is  6  oz.  13  dwt.  8  gr.  ? 

22.  What  part  of  1  gal.  is  3  qt.  1  pt.  3  gi.  ? 

23.  What  part  of  1  Cd.  ft.  is  10  cu.  ft.  1382|  cu.  in.  ? 

14:5.    To  find  a  Number  from  a  Fraetional  Part  of  it. 

1.  5861  =  ^  of  what  number? 

First  Solution.    5861  =  ^  of  7  times  5861,  which  is  41027. 
Second  Solution.    If  5861  =  f  of  some  number,  7,  or  the  number 
itself,  must  equal  7  times  5861,  which  is  41027. 
Proof.     I  of  41027  =  5861. 
Of  what  number  — 

2.  Does  3498  =^? 

3.  Does  59387  =  ^  ? 

4.  Does^^l:.:^^? 
8.   3476  =  f  of  what  number? 

Solution.  —  If  3476  =  f  of  some  number,  g  of  that  number  must  be 
^  of  3476,  which  is  -^V"^)  and  f ,  or  the  number,  must  be  9  times  thia 
result,  and  may  be  expressed  by  multiplying  the  numerator  by  9. 
Hence  we  have  the  following  written  work :  — 

869 

3476  =  g  of  ^ =  9  ''^  ~2~  "=  9  ""^  ^^^^^ 

=  Answer. 

Proof     I  of  3910  J    =3476. 


5.  Does  Mi  =  i? 

6.  Does  58.46  =  .1  ? 

7.  Does  327.93  =  .0001  ? 


FRACTIONS.  197 

9.     tf  =  M  of  what  number  ? 

Solution.  —  If  3-f  =  ff  of  some  number,  -55-  of  that  number  must 
equal  ^j-g-  of  ^f^,  which  may  be  expressed  by  making  35  a  factor  of  the 
denominator,  and  f  |,  or  the  number,  must  be  36  times  the  last  product, 
which  may  be  expressed  by  making  36  a  factor  of  the  numerator. 
This  gives  the  following  written  work :  — 

7  2 

— -  z=z  Ans. 


Hy.W~  15 

3  5 

Proof.  —  See  if  f  |  of  \%  is  equal  to  |f . 
Of  what  number  — 


10.  Does  5496  =:f? 

11.  Does  23584  =  ff  ? 

12.  Does  16875  =  if  ? 


13.  Does  f  =  ?  ? 

14.  Does -59^  =  ^3^  ? 

15.  Does  tf  =  ^^  ? 


16.  8372  —  .07  of  what  number? 

Solution.  —  Since  8372  =  .07  of  some  number,  .01  of  that  number 
must  equal  -f  of  8372,  which  is  1196,  and  \%%,  or  the  number,  must 
equal  100  times  this  result,  which,  found  by  removing  the  point  two 
places  towards  the  right,  is  119600. 

Of  what  number  — 

17.  Does  5987  =  .9?         I     19.   Does  79.84  =  .004  ? 

18.  Does  2.475  =  .03  ?       |     20.    Does  .58674  =  .0011'$ 

140.    Practical  Problems. 

1.  If  f  of  a  yard  of  cloth  cost  $3.74,  how  much  will  1 
yard  cost  ? 

Solution.  —  If  f  of  a  yard  of  cloth  cost  $3.74,  ^  of  a  yard  will  cosi 
^  of  $3.74,  which  is  $1.87,  and  f,  or  a  yard,  will  cost  3  times  $1.87, 
which  is  $5.61  =  Ans. 

The  work  may  be  written  in  either  of  the  following  forms  :  — 
First  Form. 

1.87 
i^M^  X  3 


-  $5.61 


17 


198  TRACTIONS. 

Second  Form. 

2  )  $3.74 
$1.87 


$5.61 

Proof  —  If  1  yard  of  cloth  costs  $5.61,  -j  of  a  yard  will  cost  -J-  of 
$5.61,  which  is  $1.87,  and  f  of  a  yard  will  cost  two  times  $1  87,  which 
is  $3.74. 

2.  If  ^  of  an  acre  of  land  cost  $8.54,  what  will  an  acre  cost  r 

3.  If  f  of  a  vessel  cost  $17645,  what  will  the  vessel  cost  ? 

4.  If  -^j  of  a  cask  of  oil  is  143  gallons,  how  many  gallons 
are  there  in  the  cask  ? 

5.  If  .09  of  a  lot  is  worth  $594.72,  how  much  is  'the  lot 
worth  ? 

6.  If  .13  of  a  cargo  of  coal  is  worth  $213.46,  how  much 
is  the  cargo  worth  ? 

7.  If  f  of  a  pound  of  tea  is  worth  f  ^  of  a  dollar,  what  is 
a  pound  of  tea  worth  ? 

Solution.  —  Since  the  answer  is  to  be  in  dollars,  we  write  f  f  of  a 
dollar  as  the  number  to  be  operated  upon.  If  f  of  a  pound  of  tea  cost 
so  much,  y  of  a  pound  will  cost  ^  of  this,  which  may  be  expressed  by 
making  5  a  factor  of  the  denominator ;  and  ^,  or  a  pound,  will  cost  7 
times  this  result,  which  may  be  expressed  by  making  7  a  factor  of  the 
numerator. 

This  gives  the  following  written  work  .  — 

5 

^^$  X  7  _  >,35  _        8    _ 

hi  X  $~~  hi  -  ^21  -  ^    "^^^ 

8.  If  If  of  a  bale  of  cotton  weighs  |f  of  a  ton,  how 
much  will  the  bale  weigh  ? 

9.  Bought  §J  of  a  gallon  of  oil  for  f ^  of  a  dollar.  How 
much  would  a  gallon  have  cost  at  the  same  rate  ? 

10.  If  f  of  a  yard  of  silk  cost  $1,572,  what  will  f  ^  of  a 
yard  cost  ? 

Solution.  —  Since  the  answer  is  to  be  in  dollars,  we  first  write  $1,572 
as  the  number  to  be  operated  upon.  If  f  of  a  yard  cost  $1,572,  i^  of  a 
yard  will  cost  ^  of  $1,572,  which  may  be  expressed  by  writin;^  g  undcx. 


i 


FRACTIONS.  199 

$1,572  as  a  denominator,  and  f ,  or  a  yard,  will  cost  9  times  this  result, 
expressed  by  making  9  a  factor  of  the  numerator.  If  1  yard  costs  so 
much,  It^  of  a  yard  will  cost  §y  of  this,  found  by  making  20  a  factor 
of  the  numerator,  and  27  a  factor  of  the  denominator.  This  gives  the 
following  written  work  :  — 

.131 

.^^^  10 

Note. —  Questions  like  the  above  can  always  be  resolved  mto  two 
or  more  simple  ones.  Thus  :  If  f  of  a  yard  of  silk  costs  $1,572,  what 
will  1  yard  cost  1  If  1  yard  costs  the  last  result,  what  will  f ^  of  a 
yard  cost  1 

11.  How  much  will  -Jj  of  a  cord  of  wood  cost,  if  ^  of  a 
cord  cost  $3.85  ? 

12.  If  a  ship  sails  157  miles  in  -^^  of  a  day,  how  many- 
miles  will  she  sail  in  8}  days  ? 

13.  A  man  sold  8|  tons  of  iron  for  $384,  and  afterwards 
sold  9f  tons  at  the  same  rate.  How  much  did  he  receive  for 
the  last  lot  ? 

Suggestion.  —  Reduce  the  mixed  numbers  to  improper  fractions. 
The  question  then  will  read,  —  A  man  sold  -j-  of  a  ton  of  iron  for 
$384,  and  afterwards  sold  ^^  of  a  ton,  &c. 

14.  How  much  must  be  paid  for  5^  acres  of  land,  when 
$868  are  paid  for  2f  acres  ? 

15.  If  6|  barrels  of  flour  cost  $38.25,  how  much  will  14| 
barrels  cost  ? 

16.  If  f  of  a  dollar  will  purchase  7^^  lb.  of  coffee,  how 
many  pounds  can  be  purchased  for  $5  J  ? 

Solution.  —  Since  the  answer  is  to  be  in  pounds,  we  write  7x^f ,  or 
YX,  as  the  number  to  be  operated  on.  If  |-  of  a  dollar  will  pm-chase  xt 
lb.  of  coffee,  ^  of  a  dollar  will  purchase  y  of  xT  lb.,  and  f,  or  a  dollar, 
will  purchase  8  times  this  result,  and  $5^,  or  $W  will  purchase  -V"  of 
this  last  result.    The  work,  then,  would  be  written  thus  :  — 

2         3 
..80x$X^i      480       ^^7,^ 


900  FRACTIONS. 

17.  If  12f  yards  of  calico  are  given  for  17^  yards  of 
sheeting,  how  many  yards  of  calico  must  be  given  for  25y\j 
yards  of  sheeting  ? 

Suggestion.  —  Observe  that  as  the  answer  is  to  be  yards  of  calico^  the 
number  of  yards  of  calico  is  that  to  be  operated  on. 

1 8.  If  ly^y  tons  of  hay  cost  $22 J,  how  much  will  9f  tons 
cost  ? 

19.  If  13^  lb.  coffee  are  worth  as  much  as  3f  lb.  of  tea, 
how  many  pounds  of  coffee  are  worth  as  much  as  7^  lb.  of 
tea? 

20.  If  8^  yards  of  silk  are  worth  as  much  as  24|  yards 
of  gingham,  how  many  yards  of  gingham  can  be  obtained  for 
57^  yards  of  silk  ? 

21.  When  3^^  yards  of  silk,  worth  $1|^  per  yard,  are  given 
for  If  yards  of  broadcloth,  how  many  dollars  should  be  given 
for  1  \^  yards  of  broadcloth  ? 

14:7.    Division  hy  Fractions, 

(a.)  Since  l  =  |=:|=:t  =  |,  &c.,  it  follows  that  there 
must  be  two  times  as  many  halves,  three  times  as  many 
thirds,  four  times  as  many  fourths,  &c.,  as  there  are  ones  in 
any  number. 

Or,  which  is  the  same  thing,  — 

(b.)  Since  l=z2  X  h  =  ^  X  h  =  "^  X  h  &^c.,  it  follows 
that  there  must  be  twice  as  many  times  ^,  three  times  as 
many  times  ^,  four  times  as  many  times  ^,  five  times  as 
many  times  ■^,  &c.,  as  there  are  times  1  in  any  number. 

(c.)  But  the  quotient  of  a  number  divided  by  1  equals  the 
number  itself;  hence,  the  quotient  of  a  number  divided  by  ^ 
must  equal  twice  the  number ;  divided  by  ^  must  equal  three 
times  the  number  ;  by  ^,  four  times  the  number ;  &c. 

Note.  —  This  is  but  an  application  of  the  principle,  that  if  one 
number  contains  another  a  certain  number  of  times,  it  will  contain  half 
'.hat  number  twice  as  many  times  ;  ^  of  it  three  times  as  many  times 
^  of  it  four  times  as  many  times  ;  &c. 

For  example,  — 

24  ^-  12  =  2  ;  24  -fr-  i  of  12,  or  6,  =  2  X  2,  or  4. 


FRACTIONS. 


201 


24  -J-  J  of  12,  or  4,  =  3  X  2,  or  6  ; 

24  -^  :^  of  12,  or  3,  =  4  X  2,  or  8. 

In  like  manner,  — 

1  -T-  1  =  1 ;  1  -5-  ^  of  1.  or  ^,  =  2  X  1,  or  2. 

1  -^  ^  of  1,  or  ^,  =  3  times  1,  or  3  ; 

1  -^iof  l,ori,  =  4  X  1,  or  4. 

In  like  manner,  — 

7  -»-  1  =  7  ;  7  -T-  1  of  1,  or  i",  =  2  X  7,  or  14. 

7  -^  i  of  1,  or  i,  =  5  X   7,  or  35 ; 

7  -f-  ^  of  1,  or  ^,  =  9  X  7,  or  63. 

1.   What  is  the  quotient  of  9  -^  J  ?  * 

First  Solution.    9  divided  by  1  =  9 ;  hence,  divided  by  ^,  it  must 
equal  4  times  9,  or  36. 

Second  Solution.     Since  9  contains 
times  9  times,  or  36  times. 


9  times,  it  must  contain  ^,  4 


What  is  the  quotient  — 

2.  Of  4  -f-  ^  ?  * 

3.  Of  311  -^-^? 

4.  Of  287  -^  i  ? 

5.  Of  347 -^^V? 

6.  Of  347  4-  -1  ? 


7.  Of  4.736  -^  .01  ? 

8.  Of  .75  -^  .0001  ? 

9.  Of  9000.  ~  .01  ? 

10.  Of  .0007  -^  .001  ? 

11.  Of  864 -f- .0001  ? 


12.   What  is  the  quotient  of  ^  -^-  ^  ? 

First  Solution.    |-  divided  by  1  =  |-,  and  divided  by  i^  must  equal  6 
times  I",  which,  by  cancelling,  =  3  times  \  =  ~^,  =  b^. 

Second  Solution.    Since  ^  contains  1,  |-  times,  it  must  contain  ^,  6 
times  ^  times,  =  3  times  |^  =  -^  =  5;|-  times. 

Bv  either  form  of  solution,  the  work  may  be  written  thus  :  — 

3 


7.1       7 

X 

0       21          1 

8    •   6~ 

What  is  the  quotient  — 

13.  Of  ^-f-^? 

14.  Of  1  -^T2•? 

15.  Of^f-^-jJ^? 

4 

-4-^4 

16.  Ofit^^V? 

17.  OfHI-^rV? 

18.  Off^/^-^^J^? 

*  These  questions  are  really  equivalent  to,  "  9  =  how  many  fourths  ? ' 
4  =  how  many  thirds  ?  "     (See  126,  16th  Example.) 


202  FRACTIONS. 

19.  What  is  the  quotient  of  Jf  -i-  f  ? 

First  Solution.  Jf  divided  by  1  =  ^f ,  and  by  ^  must  equal  9  timea 
^f ;  if  it  contains  ^  so  many  times,  it  must  contain  f  only  ^  as  many 
times.*    Hence,  — 

2  3 

16  ^  8_£0_x_0__6 

21    •   9  -  j^i:  X  $  ~"  T 

7 

Second  Solution,    ^f  divided  by  1  equals  ^f ,  and  divided  by  g-  mu 
equal  9  times  ^f ;  if  the  quotient  by  g-  equals  so  much,  the  quotient  by 
f  must  equal  ^  of  this  result.*    Hence,  — 

2        3 

16       8       a:0  X  0       6       .    . 

TTT  -7-  t;  ==  ;7-^ zl  =  ;^>  as  beiore. 

21    •    9       ^t  X  $       ^     ' 

7 
What  is  the  quotient  — 

20.  Of  ^  -^  /j  ? 

21.  Of  ^l-it? 

22.  Of  8ft  -4-  i  ? 

23.  Of2§|-v--3^? 

24.  OfHf-Mf*? 


25.  Off^f-^H? 

26.  Of  824 -M? 

27.  Of  617  -^T^?, 

28.  Of  675  -^  i\  ? 


29.   What  is  the  quotient  of  675  ~  .9  ? 
The  work  may  be  written  thus  :  — 

a  =  675  =  dividend 
10  X  a  =  b  =  6750  =  quotient  by  .1 

^  of  b  =  750  =  quotient  by  .9 

What  is  the  quotient  — 


30.  Of  8.47  -f-  .07  ? 
81.  Of  75.  -^  .05,? 
32.  Of  75.  ~  .005  ? 


33.  Of  .75  -^-  .05  ? 

34.  Of  .075  ^  .005  ? 

35.  Of  .00084  H-  .0012? 


*  In  accordance  with  the  principle,  that  if  a  number  contains  another  a 
certain  number  of  times,  it  will  contain  8  times  that  number  only  ^  as 
many  times  ;  12  times  the  number  only  T2"  as  many  times,  &c. 
Thus,  72  -^  2  =  36,  and  72  -f  6  times  2,  or  12,  =  i  of  36,  or  6. 
48  -^  3  =  16,  and  48  -|-  8  times  3,  or  24,  =  i  of  16,  or  2. 
7  -J  ^  =  66,  and  7  -f  6  times  i,  or  f ,  =  i  of  56,  or  -"V-  =  ll-^. 
t  -^  tV  =  ■¥^,  and  F  -^  7  times  A,  or  /a:,  =  I  of  ■%°-,  or  ti 
t  Reduce  to  an  improper  fraction. 


FRACTIONS.  203 

14:8,    Process  of  Division  generalized. 

(a.)  Since  the  quotient  of  any  number  divided  by  ^  r=  5 
times  the  number ;  by  ^  =:  9  times  the  number ;  by  ^g-  = 
13  times  the  number,  &c. ;  and  since  the  quotient  of  a  num- 
ber divided  by  f  =  ^  its  quotient  divided  by  -^ ;  divided  by 
I  m  ^  of  its  quotient  divided  by  ^  ;  divided  by  \^  =  -^  of 
its  quotient  divided  by  y^,  &c.,  it  follows  that  the  quotient  of 
a  number  divided  by  — 

-|  =  ^  of  5  times  the  number,  =  |^  of  the  number ; 

|.  ==  ^  of  9  times  the  number,  z=  |  of  the  number ; 

\^  =:  -^Y  of  13  times  the  number,  =z  ^  of  the  number; 

.07  =  ^  of  100  times  the  number  =  -i^^  of  the  number ; 
and,  universally,  that  — 

The  quotient  of  any  number  divided  by  a  fraction,  is  equal 
to  the  product  of  that  number  multiplied  by  the  fraction 
inverted. 

Illustrations. —  1.  To  divide  by  f ,  we  have  only  to  multiply  by  9  and 
divide  by  8. 

2.  To  divide  by  f y,  we  have  only  to  multiply  by  1 7  and  divide 
by  3. 

3.  To  divide  by  .03,  we  have  only  to  multiply  by  100  and  divide  by 
3,  or,  which  is  the  same  thing,  to  divide  by  3  and  remove  the  point  2 
places  towards  the  right. 

4.  To  divide  by  .000037,  we  have  only  to  multiply  by  1000000,  and 
divide  by  37,  or,  which  is  the  same  thing,  to  divide  by  37,  and  rf 

the  point  6  places  towards  the  right. 

1.  What  is  the  quotient  of  A  "T"  t?  ? 

5     .    10__    g  X  13  _  13 

^''''  Ti  "^  T3  "~  11  X  ^0  ~  2^ 

2 

2.  What  is  the  quotient  of  520.6  -^  .011  ? 

^,i5.__The  quotient  of  520.6  -r-  .011  maybe  obtained  by  dividing 
B20.6  by  11  and  removing  the  poin^  *hree  places  towards  the  right, 
thus :  — 

First  Form.  Second  Form. 

.011  )  520.6  Oil  )  520.600 

47327^  47327^3^ 


204  FRACTIONS. 

Note.  —  The  second  form  of  writing  the  work  differs  from  the  first 
only  in  this,  —  that  in  it  as  many  zeros  are  annexed  to  the  dividend  as 
would  be  necessary  were  the  point  actually  changed  before  performing 
the  division.  Great  care  is  necessary,  by  either  of  these  forms,  to  insure 
that  the  point  is  placed  correctly  in  the  quotient ;  and  if  in  any  case 
there  is  a  doubt  as  to  its  true  position,  the  work  should  be  written  in 
full,  as  in  the  model  given  after  example  29th,  147. 

3.   What  is  the  quotient  of  f  of  J  of  ^  -r  f  of  ^f  of  f  ^  ? 

Solution.  — 


4 
5 

of^oi 

,  8 
11    * 

8    .12 
"9°^  25 

of 

21 

22 

= 

(^ 

X  7 
X  9 

X 
X 

8 
11 

5 

2 

8  X 

12  X 
2b  X 

21\* 

22/    - 

4 

X  ^ 

X 

0 

X 

0  X 

:  ^$ 

X 

^^ 

9  X 

X  0 

X 

:iX 

X 

$  X 

:  x^ 

X 

^t 

10 

,1 

3 

3 

9  ~ 

^^9 

Note.  —  The  more  fhll  and  analytical  explanation  would  be  the 
following :  yy  is  the  number  to  be  operated  upon.  ^  of  -yr  may  be 
expressed  by  making  7  a  factor  of  the  numerator,  and  9  a  factor  of 
the  denominator.  ^  of  this  may  be  expressed  by  making  4  a  factor  of 
the  numerator,  and  5  a  factor  of  the  denominator.  The  quotient  of 
this  quantity  divided  by  f  g-  will  be  f  f  of  it,  and  may  be  expressed  by 
making  22  a  factor  of  the  numerator,  and  21  a  factor  of  the  denomina- 
tor. If  f^  is  contained  so  many  times,  ^f  of  f  ^  must  be  contained 
^^  as  many  times,  expressed  by  making  25  a  factor  of  the  numerator, 
and  12  a  factor  of  the  denominator.  If  this  divisor  (^§  of  |^)  is  con- 
tained so  many  times,  f  of  it  must  be  contained  f  as  many  times, 
expressed  by  making  9  a  factor  of  the  numerator,  and  8  a  factor  of  the 
denominator.    Hence,  — 


4       7        8     .    8    .12       21 
6"^9"^n-9°^25°^2^  = 
2         5 

$x;^x^x^^x^^x0      10 

4 

nx^x$x^txXfix$'-9~ 

3         3 

*  The  equation  in  the  parenthesis  may  be  omitted  in  practical  opera- 
tions 


FRACTIONS. 


205 


7.  Of  .004  -f-  25  ? 

8.  Of  .067  -^  .02  ? 

9.  Of  3287  -^  .0004  ? 


What  is  the  quotient  — 

4.  Of  13f  -^  8^  ? 

5.  Of  llf -^  4^? 

6.  Of  16f -f-  28|? 

10.  Of  f  of  ii-  of  If  -^  ^1  of  ^Vf  of  ^«^? 

11.  Of  f  of  -y-  of  f^  -h  I-  of  y9^  of  6^  ? 

12.  Of  ^  of  ^  of  J  of  -i  of  ^  of  I  -^  i  of  i  of  T\y  of  ^\  ? 

13.  Of  f  of  f  of  ^  of  f  -^-  ^7^  of  if  of  li  ? 


14:0.     Complex  Fractions. 

(a.)   A  COMPLEX  FRACTION  is  One  having  a  fraction  in 

7     2.*     § 
either  numerator  or  denominator,  or  in  both ;  as,  — ,  -?,  -^. 

3f'  4^'  7| 

Note.  —  Complex  fractions  are  usually  considered  as  expressions  of 

unexecuted  divisions,  and  are  read  accordingly.     Thus,  — 

(J).)    To  show  their  similarity  to  other  fractions,  we  may 
explain  them  thus :  — 

—  =  7  parts  of  such  kind  that  3§-  of  them  would  equal  a  unit 

(c.)    Complex  fractions  can  be  reduced  to  simple  fractions 

^y  the  ordinary  process  of  division. 

54- 
1.   Reduce  ^  to  a  simple  fraction. 


Solution.    ^  = 

36=" 

7 

«=    ^    37  _  36  X    4 
•     4          7    X  37 

_  144 
~  259 

, 

Reduce  each  of  the  following  to  simple 

fractions. 

«■* 

5.     «^. 
12i 

8.  li. 

8 

.a 

6.   5i. 

4? 

9.     !i. 

11 

'•ft 

7      13* 
Hi 

10.    4? 
5« 

*  By  reducing  5i  to  sevenths,  and  9^  to  fourths. 
18 


206 


FRACTIONS. 


(d.)    Complex  fractions  may  often  be  reduced  to  simple 
ones,  by  reducing  them  to  their  lowest  terms,  — 

3  3  

Thus :  Dividing  both  terms  of  —  by  1 3-  gives  ^  —  §.    Dividing 

if  l| 

both  terms  of  -^  by  if  gives  —  =  ^. 
5  5 

Reduce  each  of  the  following  in  the  same  manner :  — 


11.  -a 

5 

-h 

17.  -21 
16f 

12.  If. 

6? 

18.  21. 
9* 

13.  7*. 

-•  3- 

19.  1!. 

{e.)    Complex  fractions   may  also  be   reduced  to  simple 

ones,  by  multiplying  both  numerator  and  denominator  by  such 

a  number  as  will  give  a  whole  number  in  place  of  each. 

42 
20.   Reduce  — ^  to  a  simple  fraction. 
10^ 

Solution.  —  If  4§  be  multiplied  by  3,  or  some  multiple  of  3,  and  10^ 
be  multiplied  by  2,  or  some  multiple  of  2,  the  result  will  in  each  ease  be 

a  whole  number.    Hence,  if  both  terms  of  the  fraction  — ^  be  multi- 

plied  by  some  multiple  of  both  2  and  3,  the  resulting  fraction  will  be  a 

simple  one.    Multiplying  by  6  gives  — —  =  —  =:-. 

1 0^         63         9 

In  the  same  way  reduce  each  of  the  following  complex 
fractions  to  simple  ones :  — 


- 1 

24.  7*. 
13^ 

27.  l«i 
83i 

-  3- 

25.  "^^l 
627i 

28.  >«3| 
847 

-  g- 

26.  "«*. 
64? 

29.  *7i 
281 

FRACTIONS.  207 

150*    Other  Changes  in  the  Terms  of  a  Fraction. 

1.  Reduce  J  to  an  equivalent  fraction,  having   6  for   a 
numerator. 

Solution.  —  Observing  that  the  proposed  numerator,  6,  is  two  times  the 

3  3X2 
^tven  numerator,  3,  we  multiply  both  terms  by  2,  which  gives  ■-  = 

4  4  X   -a 

:=  -,  or  we  niay  at  once  write  -  =  — 
8  4         8 

2.  Reduce  f  to  an  equivalent  fraction,  having  10  for  its 
numerator. 

10        5 
^Zirfwn.  —  Observing  that  the  proposed  numerator,  10,  is  —,  or  -,  of 

5  . 

the  given  numerator,  8,  we  multiply  both  terms  by  -,  or  by  1^^,  which 
,  4 

8        8  X  Ix        10  8        10 

gives  -  = =  — ,  or  -  =  — . 

^         9       9X11.       14        9        11^ 

3.  Reduce  ^  to  an  equivalent  fraction,  having  10  for  its 
numerator. 

4.  Reduce  y%  to  an  equivalent  fraction,  having  9  for  its 
numerator. 

5.  Reduce  ^^  to  an  equivalent  fraction,  having  6  for  its 
numerator. 

6.  Reduce  f  to  twenty-firsts. 

Solution.  —  Observing  that  the  proposed  denominator,  21,  is  three 

times  the  given  denominator,  7,  we  have  only  to  multiply  both  terms 

2  2X36 

by  3,  which  gives  -  =  =  — ,  or,  omitting  to  write  the  inter- 

"^    '  ^         7         7   X  3        21'      '  ^ 

mediate  work,  we  have  -  =  — 
7       21 

Note.  —  The    same    result    might  have   been    obtained   thus :  — 
21   ,  2    ,  ,  2    ^  21     1    ^  21  .     3 

1  =  — ,  hence  r  of  1  must  equal  y  of  —;  7  ®'  57  ^^  5T»  ^"^  '^®  times 

—  =  ^ )    l>it|  ii  practice,  the  first  form  will  usually  be  found  most 
convenient. 

7.  Reduce  f  to  fifteenths. 

15 
Solution.  —  Observing  that  the  proposed  denominator,  1 5,  is  —  of  the 

15  1 

given  denominator,  7,  we  have  only  to  multiply  both  terms  by  — ,  or  2-. 

„  2       2  X  2^       4f         2        4f 

^^°^^'T  =  rX^  =  r5'«^7  =  T5" 


208  FRACTIONS. 

8.   Reduce  |-  to  halves. 

2 

Solution.  —  Observing  that  the  proposed  denominator,  2,  is  -  of  the 

2 

given   denominator,  9,  we  have  only  to  multiply  both  terms  by  - 

Hence  5-5Xf_li        5_li 

Reduce  — 
9.     §  to  ninths. 

10.  ^2-  to  fourths. 

11.  T®r  to  forty-fifths. 

12.  -I  to  twelfths. 

13.  I  to  forty-ninths. 


14.  f  to  thirtieths. 

15.  ^  to  twenty-sevenths. 

16.  -^T  to  ninths. 

17.  -f^  to  sixty-fifths. 

18.  -j^-j-  to  twenty-seconds. 


1^1.    Reduction  to  a  Common  Denominator. 

{a.)   Fractions  having  their  denominators  alike  are  said  to 
have  a  common  denominator. 


Thus,  4,  3.,  6,  and  X  have  a  common  denominator,  but  A  and  _ 


T'  T'  T' T '  ""-  T ~ 

have  not. 

(h.)  In  reducing  fractions  having  different  denominators  to 
a  common  denominator,  (i.  e.,  to  equivalent  ones  having  the 
same  denominator,)  we  first  select  a  convenient  number  for 
the  common  denominator,  and  then  make  the  reductions  as  in 
the  last  article. 

(c.)  As  far  as  the  denominator  is  concerned,  one  number 
may  as  well  be  selected  for  a  common  denominator  as  another i 
but  unless  the  number  selected  is  a  common  multiple  of  all 
the  given  denominators,  one  or  more  of  the  resulting  numera- 
tors will  be  likely  to  contain  a  fraction.* 

{d.)  To  avoid  such  an  inconvenience,  and  at  the  same  time 
to  avoid  as  far  as  possible  the  use  of  large  numbers,  it  will 
usually  be  best  to  select  the  least  common  multiple  of  tho 
given  denominators  for  a  common  denominator. 

1.    Reduce  |,  j^,  J,  and  ^f  to  a  common  denominator. 

*  For  it  will  be  the  product  of  a  whole  number  multiplied  by  a  frac- 
ticual  quantity. 


i 


FRACTIONS.  209 

Solution.  —  We  select  72  for  the  least  common  denominator,  because 
it  is  the  least  common  multiple  of  the  given  denominators. 

Then,  since  72  =  9  times  8,  we  multiply  both  terms  of  the  fraction 
I  by  9,  which  gives  |-  =  ff.  Since  72  =  6  times  12,  we  multiply  both 
terms  of  the  fraction  -f*^  by  6,  which  gives,  &c. 

Hence,  i  =  M5  l^  =  f  ^  5  t  =  M;  and  ^  =  ?|. 

(<?.)  Many  adopt  it  as  a  general  rule,  to  select  the  product 
of  the  given  denominators  for  a  common  denominator ;  but  it 
usually  involves  larger  numbers  than  the  preceding  method, 
and  is  hence  much  less  convenient.  The  following  illus- 
trates it. 

Solution  to  preceding  Example.  —  The  product  of  8,  9,  12,  and  24,  the 
given  denominators,  is  20736,  which  we  select  for  the  common  denomi- 
nator. To  get  this,  we  multiplied  8,  the  denominator  of  |-,  by  9  X  12 
X  24,  and  therefore  we  multiply  the  numerator,  7,  by  the  same  num- 
bers, which  gives  f  =  if yff .  To  obtain  20736,  we  multiplied  12,  the 
denominator  of  i^,  by  8  X  9  X  24,  the  product  of  the  other  denomi- 
nators, and  therefore  we  multiply  the  numerator,  5,  by  the  same  numbers, 
which  gives  x^g  =  /uyW-  To  obtain  20736,  we  multiplied  9,  the  de- 
nominator of  I",  by,  &c. 

(/•)  When  any  of  the  fractions  to  be  reduced  are  com- 
pound or  complex,  they  must  first  be  reduced  to  simple  ones, 
and  the  simple  fractions  should  be  reduced  to  their  lowest 
terms,  except  when  to  do  it  would  increase  the  labor  of  redu- 
cing to  a  common  denominator. 

{g.)  Reduce  the  fractions  in  each  of  the  following  exam- 
ples to  a  common  denominator :  — 

o      4    2    5        ,7 
2.     _,  _,  _,  and  — 
5    3    6  15 

Q      3    5    1    1        ,4 
6.     -,  _,  -,  -,  and  — . 
7    9    2    6  21 

.      19  23  41    37        ,  49 
4.     — -  — ,  — , ,  ana . 

24  36  72  108  216 

.      24   7    14    37         ,  44 
o.     — ,  _,  — , and  — . 

25  9    15    225  75 

.11111        ,1 
D.     -,  -,  — ,  — ,  — ,  and  — . 
2    3    4   5'  6  12 

18* 


210  FRACTIONS. 

7.  _,  and  -. 
f  3'  5  2 

Q      13    13    13        ,  13 

8.  — ,  — ,  — ,  and  -^. 
14    21    28  25 

^      2n35.8         ,2.  12 

9.  _  of  -,  -  of  — ,  and  -  of  -— . 
3        4'  6        15'  9         16 

10.  ?il,  L  of  7^,  and  L, 
8f      11  ^  5^ 

11.  ?i,  5,  ?l,  and  I. 

4|  6  ^      9 

12.  1  of  15,  A,  ^-L,  ?*,  and  1|. 
5        16   32    16^    7^  32 

13.  1,  1?,  1?,  ?  of  !i,  M,  and  L^l 
11    16   22^  9        22     9  100 


15S.    Addition  and  Subtraction  of  Fractions. 

(a.)  Fractions,  like  other  numbers,  must  be  of  the  same 
denomination,  in  order  to  be  added  or  subtracted. 

(h.)  To  be  of  the  same  denomination,  they  must  (131, 
e.)  be  fractions  of  the  same  unit,  and  also  have  a  common 
denominator. 

J  -\-i  cannot  be  added  in  their  present  form,  any  more  than  can  3 
pounds  and  4  ounces. 

y  of  a  yard  and  f  of  an  inch  cannot  be  added  in  their  present  form, 
any  more  than  can  3  yards  and  3  inches. 

(c.)  When  abstract  fractions  are  given,  (i.  e.,  fractions  of 
the  abstract  unit,  as  f,  ^,)  they  may,  if  simple,  be  reduced  at 
once  to  a  common  denominator ;  but  compound  and  complex 
fractions  must  be  reduced  to  simple  ones,  and  fractional  parts 
of  denominate  numbers,  as  ^  of  a  yard,  ^  of  a  foot,  must  be 
reduced  to  fractions  of  the  same  unit,  before  they  are  reduced 
to  a  common  denominator. 

1.    What  is  the  sum  of  ^  +  f  +  §  +  1^  +  f  ? 

Solution.  —  By  reducing  to  a  common  denominator,  we  have  — 


FRACTIONS.  211 

752113^2120162218^ 
8  """  6  "^  3  "^  12  "*"  4         24  "^  24  '*■  24  "^  24  "^  24 
21  +  20  +  16  4-  22  +  18  _       1 
24  ^^' 

2.   What  is  the  sum  of  87^  +  13|  +  27|  +  ^1-^^  + 

48tV? 

Solution.  —  The  sum  of  the  whole  numbers  is  212.     Reducing  the 

fractions  to  a  common  denominator,  we  have  i"  +  f  -f"  F  +  tV  +  T2" 

=  H  +  §1  +  if  +  ^^  +  M  =  2f  f ,  which  added  to  212  =  214t f  • 

The  following  arrangement  of  the  numbers  shows  the  resemblance 

of  the  work  to  compound  addition :  — 

Ones,  36ths. 
87^  =  87  18 
\S^  =  13  30 
27f  =  27  16 
37i^  =  37  10 
48/2-  =  48       21 


214       23  =  214|^. 

3.  What  is  the  sum  of  -^-^  of  a  ton  -|-  |  of  a  quarter  ? 
Solution.  —  By  reducing  the  -[^  of  a  ton  to  cwt.  and  qrs.,  and  adding 

the  ^  of  a  quarter,  we  should  have  the  following  written  work :  — 

T^2-  T.  -f  t  qr.  =  11  cwt.  2f  qr.  +  |  qr.  =  11  cwt.  3^-  qr.  =  11  cwt. 
3qr.  12  lb.  8  oz. 

4.  What  is  the  value  of  343|  — 138ff  ? 

Solution.  —  By  reducing  the  fractions  to  a  common  denominator,  we 
have  343|  —  138f  J  =  343||  —  138f f  =  204|f. 
Or  the  work  may  be  written  thus  :  — 
Ones.    72ds. 
343|     =343       16  =  Min. 
138ff  =  138       69  =  Sub. 

204       19  =  204^9  =  Rem. 

(d.)  When  several  fractions  are  to  be  added,  it  will  often- 
times be  a  saving  of  labor  to  consider  at  first  only  two  of 
them,  and  then  a  third  with  the  sum  of  these  two,  and  then  a 
fourth  with  the  sum  of  these,  and  so  on,  till  all  are  added. 
Care  should  be  taken  to  couple  tl:em  in  such  a  way  as  to 
make  the  reductions  easy. 

Thus,  in  the  24th  example  below,  t  -f  f  =  1^- ;  1^  -f  ^  ==■  9  ;  and 
2  +  f  =  2f  =  Ans. 


212 


FRACTIONS. 


Perform  the  following  examples  : 
1        1 


5      ^4-   ^ 
^'     8  +  i0* 


6.     '-  +  ' 


9 

7.  !  +  A. 
8^  12 

8.  H  +  21 

9.  3|  +  lf 

19.     8li--4l5. 
17i         11 


20. 


21. 


18       ^ 
25  "^  15' 
13__1 
18       4* 
27___9^ 
30       10* 


22.     —^ 


23.    !^ 


24 


10.     -_ 
2 

3 

i" 

7 

8" 

13.    3^ 


11 


12 


2f 
14.     5J-~3f. 


1  .  1 
4"^5 
1__1 
4  -5 
35__35 
52       78 

5i 


15. 


16.    ±_i. 


17.    _ 


18.     2^_H 
16 


H  "^  7| 

-  +  -  +  -  +  -. 
6^4^3^2 


8f. 


25.  9J 

26.  l  +  Bi  +  7i  +  l^ 

27     !4.^_lA_i.1! 
9^8^12^36* 

1  49 

28.     9f  +  -  +  4H  +  ^' 


9 


75 


29      14-^^?  +  ? 
^^'     12  +  6^^4  +  3* 

30.     5t  +  4,e,+A  +  | 


31.     2^  +  3^ 


4i  +  5i-^. 


^^-  S''^^^4+^*+^^+^- 


17       95 
^^-     24  +  96 


4    .7    .4^    .2 
-ofgofg^of-. 


34. 


|of,^  +  |orfof4| 


of;. 


357       294       423 

853  "^  671  "^  849' 

o.      „6|     ,    2^         5 

'«•    ^-Ti  +  Tii- 


llf    •    14f        6        12^        4^ 


FRACTIONS.  .  '213 

<.7     47,   83        79         3^        8f       4A,    5       12 
lOf       2 

/»  1 

38.  -  of  a  bu.  +  -  of  a  peck. 

8  5 

39.  vT  of   a  m.  +  -  of  a  ftirlong. 

17  2 

40.  -^  of  a  £  —  Q  of  a  shilling. 

7  3  1 

41.  -  of  a  lb.  +  2  ^^  a  *^^*'*  +  9  of  a  grain. 

42.  -  of  a  bu.  +  -  of  a  pk.  —  — ^  of  a  quart. 

5  2  2 

43.  ^  of  a  yd.  +  -  of  a  qr.  -|-  -  of  a  nail. 

6  o  o 


SECTION  XI 


APPLICATIONS   OF    FOREGOING    PRINCIPLES. 

1^3.    Introductory  Note. 

In  performing  the  examples  of  this  section,  as  of  every  other,  the 
student  should  observe  that  the  solutions  and  models  of  writing  the 
work  are  designed  to  suggest  methods  of  applying  principles,  and  are 
not  to  be  regarded  as  forms  which  must  be  rigidly  followed.  They 
should  be  deviated  from  when  better  forms  can  be  discovered,  or  will 
apply.  Indeed,  a  habit  of  examining  each  question  carefully  before 
attempting  its  numerical  solution  is  one  of  the  most  valuable  a  pupil 
can  acquire.  A  distinguished  teacher  once  said,  "  If  I  were  required, 
on  peril  of  my  life,  to  perform  a  complicated  problem  in  two  minutes,  I 
would  spend  the  first  minute  in  considering  how  to  do  it." 


214  '  FRACTIONS  ' 

154:.    Miscellaneous  Problems. 

1.  How  much  will  7^  yards  of  silk  cost  at  $1 J  per  yard  ? 

2.  If  7|^  yards  of  silk  cost  $14,  how  much  will  1  yard 
cost  ? 

3.  How  many  yards  of  silk,  at  $1|-  per  yard,  can  be 
bought  for  $14? 

4.  How  much  will  a  lot  of  land  28J  rods  long  and  23^ 
rods  wide  cost,  at  $2^^^  per  acre  ? 

5.  How  many  pounds  of  coffee,  at  15^  cents  per  lb.,  can 
be  bought  for  $8.40  ? 

6.  How  many  square  feet  and  inches  are  there  in  a  board 
13^  feet  long  and  2-^  feet  wide  ? 

7.  How  many  square  feet  in  the  four  walls  of  a  room,  15^ 
feet  long,  12f  feet  wide,  and  8^  feet  high  ? 

8.  I  bought  a  cask,  containing  94^  gallons  of  oil,  at  $1,375 
per  gallon ;  f  of  it  leaked  out,  and  I  sold  the  remainder  at 
$1.50  per  gallon.     How  much  did  I  lose  by  the  transaction  ? 

9.  Mr.  Whitney  is  worth  $1473.21  more  than  Mr.  Whip- 
ple, and  Mr.  Whipple  is  worth  just  f  as  much  as  Mr.  Whit- 
ney.   How  many  dollars  is  each  of  them  worth  ? 

Suggestion.  —  If  Mr.  "Whipple  is  worth  only  f  as  much  as  Mr.  Whit- 
ney, the  difference  between  the  values  of  their  estates  must  equal  ^ 
of  Mr.  Whitney's  estate. 

10.  What  is  the  value  of  |  of  5i  of  ||  -^  ||  of  \  of 

7        9§        11         5^        4 

?I?tof^? 
6A       If 

11.  What  is  the  value  0^5  +  3  —  iq  +  ^^ +  3  —  5  + 

-  +  -? 
4^15 

12.  If  8^  yards  of  broadcloth  cost  $29f,  how  much  will 
10|  yards  cost  ? 

13.  If  8^  yards  of  broadcloth  can  be  purchased  for  $29f, 
how  many  yards  can  be  purchased  for  $35^  ? 

14.  If  10|  yards  of  broadcloth  cost  $35^,  how  much  will 
%^  yards  cost  ? 


1 


FRACTIONS.  215 

15.  If  $35^  are  paid  for  10|  yards  of  broadcloth,  for  how 
many  yards  should  $29f  be  paid  ? 

16.  If  -f^  of  a  ton  of  hay  costs  $17.50,  how  much  will 
two  loads  cost,  one  weighing  f  of  a  ton,  and  the  other  ^|  of 
a  ton? 

17.  If  22  horses  eat  41^  bushels  of  grain  in  lOJ-  days, 
how  many  bushels  will  21  horses  eat  in  9^  days  ? 

Solution.  —  As  the  answer  is  to  be  in  bushels  of  grain,  we  write  41^, 
or  ^^^^  as  the  number  to  be  operated  on.  If  22  horses  eat  -^  j-^  bushels, 
1  horse  will  eat  sV  as  much  in  the  same  time,  which  may  be  expressed 

by  making  22  a  factor  of  the  denominator :  thus, '■ .    If  1  horse 

•^  '  4    X  22 

will  eat  this  quantity  in  10^,  or  ^-,  days,  in  ^  day  he  will  eat  2^-  of 

this,  and  in  2  half-days,  or  a  day,  he  will  eat  twice  this  result,  expressed 

by  making  21  a  factor  of  the  denominator,  and  2  a  factor  of  the  numera- 

165  X  2 
tor;  thus,  — —  ■*    If  1  horse  eats  this  quantity,  21  horses 

will  in  the  same  time  eat  21  times  this,  expressed  by  making  21  a  factor 

165  X    2   X  21  * 

of  the  numerator ;  thus,  — —  .t    If  21  horses  eat  so  much  in 

4     X  ^^  X  ^i- 

1  day,  in  9^,  or  ^^-,  days,  they  will  eat  -^  of  this,  which  may  be  ex- 
pressed by  making  28  a  factor  of  the  numerator,  and  3  a  factor  of  the 

-  .     ^  .        165  X    2    X  21  X  28^ 
denommator ;  thus,    ^    x  22  X  21  X   3 '^ 

-  Hence  the  written  Avork  and  answer  are  as  follows  :  — 

5 

X$  7 


Note.  —  It  will  be  seen  that  in  the  solution,  each  condition  of  the 
question  was  considered  by  itself,  without  reference  to  the  other  con- 
ditions, and  that  the  problem  was  dealt  with  as  though  composed  of  the 
following  series  of  simpler  problems  :  — 

1.  If  22  horses  eat  415-  bushels  of  grain  in  a  given  time,  how  many 
bushels  would  1  horse  eat  in  the  same  time  1 

*  This  shows  how  many  bushels  1  horse  will  eat  in  1  day. 
t  This  shows  how  many  bushels  21  horses  will  eat  in  1  day. 
J  This  shows  how  many  bushels  21  horses  will  eat  in  9^  days. 


216  FRACTIONS. 

2.  If  I  horse  eats  the  quantity  ohtained  as  the  answer  to  the  last 
question  in  lO^-  days,  how  many  bushels  will  he  eat  in  1  day  ? 

3.  If  1  horse  eats  so  much  (the  answer  last  obtained)  in  a  day,  hoTT 
many  bushels  will  21  horses  eat  in  the  same  time  ? 

4.  If  21  horses  eat  so  much  (the  answer  last  obtained)  in  1  day,  how 
many  bushels  will  they  eat  in  9^  days  ? 

The  student  should  notice  that  the  question  in  each  case  has  been, 
.  "  How  many  bushels  ?  "  and  that  the  denomination  of  each  answer  has 
been  the  same  as  that  required  for  the  final  answer. 

18.  If  22  horses  can  be  kept  10^  days  on  41^  bushels  of 
j^rain,  how  many  hordes  can  be  kept  9^  days  on  35  bushels 
of  grain. 

19.  If  22  horses  can  be  kept  10^  days  on  41^  bushels  of 
grain,  how  many  days  can  21  horses  be  kept  on  35  bushels 
of  grain  ? 

20.  If  21  horses  eat  35  bushels  of  grain  in  9^  days,  how 
many  bushels  will  22  horses  eat  in  10^  days  ? 

21.  If  21  horses  can  be  kept  9^  days  on  35  bushels  of 
gi'ain,  how  many  days  can  22  horses  be  kept  on  41 1-  bushels 
of  grain  ? 

22.  If  21  horses  can  be  kept  9^  days  on  35  bushels  of 
grain,  how  many  horses  can  bie  kept  10^  days  on  41 1-  bushels 
of  grain  ? 

23.  If  7  men,  by  working  8^  hours  per  day  during  6  days 
of  the  week,  can  earn  $320f  in  5^  weeks,  how  many  dollars 
would  4  men  earn  in  4|  weeks,  by  working  9^  hours  per  day, 
and  5  days  per  week  ? 

24.  If  it  costs  $9.25  to  dig  a  ditch  47  feet  long,  1|  deep, 
and  3  feet  wide,  how  much  will  it  cost  to  dig  a  ditch  70^  feet 
long^  2^  feet  deep,  and  4^  feet  wide  ? 

25.  If,  when  flour  is  $6§  per  barrel,  a  six-cent  loaf  weighs 
15  oz.,  how  many  ounces  ought  a  ten-cent  loaf  to  weigh  when 
flour  is  $9  J  per  barrel  ? 

26.  If  a  block  of  oak,  3f  feet  long,  1  foot  wide,  and  f  of  a 
foot  thick,  weighs  120|  lb.,  how  much  will  a  block  of  pine,  8 
^eet  long,  2  feet  wide,  and  If  feet  thick  weigh,  pine  being 
only  f  as  heavy  as  oak  ? 

27.  If  9  men,  by  working  10  hours  per  day  durinir  6  days 


FRACTIONS.  217 

of  the  week,  can  in  4  weeks  dig  a  trench  450  feet  long,  3^ 
feet  wide,  and  2|  feet  deep,  how  many  men,  working  9^  hours 
per  day  during  5  days  of  the  week,  can  in  9  weeks  dig  a 
trench  539  feet  long,  6^  feet  wide,  and  2|^  feet  deep  ? 

28.  If  12  men,  by  working  9^  hours  per  day  during  5 
days  of  the  week,  can  in  9  weeks  dig  a  trench  539  feet  long, 
f)^  feet  wide,  and  2|^  feet  deep,  how  many  weeks  would  it" 
take  9  men,  working  10  hours  per  day  during  6  days  of  the 
week,  to  dig  a  trench  450  feet  long,  3^  feet  wide,  and  2|  feet 
deep  ? 

29.  If  36  lb.  of  sugar  are  worth  24  lb.  of  coffee,  and  22 
lb.  of  coffee  are  worth  55  lb.  of  rice,  how  many  pounds  of 
rice  can  be  bought  for  16  lb.  of  sugar  ? 

Preliminary  Explanation.  —  As  the  answer  is  to  be  pounds  of  rice, 
"we  nriust  write  55.  the  given  number  of  pounds  of  rice,  as  the  number 
to  be  operated  on,  and  must  take  care  to  express  the  values  throughout 
in  pounds  of  rice.    We  should  then  have  the  following 

Solution.  —  Since  22  lb.  of  coffee  are  worth  55  lb.  of  rice,  1  lb.  of 
coffee  must  be  worth  ^V  of  55  lb.  of  rice,  which  may  be  expressed  by- 
writing  22  as  a  denominator  under  the  55  ;  and  24  lb.  of  coffee  must  be 
worth  24  times  the  last  result,  which  may  be  expressed  by  making  24  a 
factor  of  the  numerator.  But  as  this  is  also  the  value  of  36  lb.  of  sugar, 
1  lb.  of  sugar  must  be  worth  -^^  of  this,  which  may  be  expressed  by 
making  36  a  factor  of  the  denominator;  and  16  lb.  of  sugar  must  be 
worth  16  times  the  last  result,  which  may  be  expressed  by  making  16  a 
factor  of  the  numerator. 

The  work  would  be  written  thus  :  — 

5        t^ 
lb.  of  rice  ^^  ii  ^^  ^  ^^   =   26f  lb.  of  rice  =  Ans, 
^  3 

Note.  —  Examples  like  the  above  are  usually  solved  by  a  process 
^called  Conjoined  Proportion  ;  but  as  it  is  much  less  simple  and  con- 
^enient  than  the  preceding,  we  omit  it. 

30.  If  40  bushels  of  potatoes  are  worth  45  bushels  of  corn, 
and  18  bushels  of  corn  are  worth  14  cwt.  of  hay,  and  35  cwt. 
of  hay  are  worth  4  barrels  of  flour,  how  many  barrels  of 
flour  are  75  bushels  of  potatoes  worth  ? 

19 


218  PRACTICE. 

31.  If  7  American  yards  are  equal  to  12  braces  at  Leg* 
horn,  and  54  braces  at  Leghorn  are  equal  to  45  braces  at 
Venice,  how  many  braces  at  Venice  are  equal  to  56  American 
yards  ? 

32.  If  John  can  earn  as  much  in  7  days  as  William  can 
earn  in  9  days,  and  William  can  earn  as  much  in  15  days  as 
Samuel  can  in  14,  and  Samuel  can  earn  as  much  in  12  days 
as  Otis  can  in  8,  and  Otis  can  earn  as  much  in  24  days  as 
Eufus  can  in  21,  how  many  days  will  it  take  John  to  earn  as 
much  as  Eufus  can  earn  in  49  days  ? 

33.  Multiply  52^  by  7|,  subtract  14f  from  the  product, 
and  divide  by  f. 

34.  A  man  gave  4^  acres  of  land,  worth  $125.37  per  acre, 
in  exchange  for  75  barrels  of  flour.  He  sold  f  of  the  flour 
at  $9 1  per  barrel,  and  the  rest  at  $8^  per  barrel.  How 
much  did  he  gain  by  the  transaction  ? 

35.  A  garrison  of  2000  men  had  bread  enough  to  allow 
each  24  oz.  a  day  for  75  days ;  but  being  besieged,  it  received 
a  reenforcement  of  1600  men.  How  many  ounces  per  day 
can  each  man  be  allowed,  in  order  that  they  may  hold  out  60 
days? 

36.  Half  of  Arthur's  money  equals  just  f  of  William's, 
and  William  has  $187.47  more  than  Arthur.  How  much 
money  has  each  ? 

1S5.    Practice. 

The  following  examples  illustrate  what  is  sometimes  called 
the  Rule  of  Practice. 

1.  How  much  will  47  yd.  3  qr.  3  n.  of  broadcloth  cost,  at 
$3,125  per  yd.  ? 

Suggestion.  —  Since  47  yd.  3  qr.  3  na.  lacks  but  1  nail  of  being  48 
yards,  its  cost  must  equal  the  cost  of  48  yards,  minus  the  cost  of  1  nail 

WRITTEK  WORK. 
B  =  $     3.125 


48  X  a  =  b  =  $150,000  =  cost  of  48  yds. 
lVofa  =  c  =  $       .195=     "     '•    1  na. 

b  —  c  =  Si 49.805  r^     "     "    47  yds.  3  qr.  3  na. 


J6  X  a  = 

=  b  = 

$4530.50 

= 

iof  a  = 

=  c  = 

43.562 

= 

^of  c  = 

=  d  = 

21.781 

= 

iof  d  = 

=  e  = 

5.445 

= 

PRACTICE.  219 

2.  How  much  will  26  A.  1  R.  25  sq.  rd.  of  land  cost,  at 
S174.25  per  acre  ? 

WRITTEN  WORK. 

a  =  $  174.25    =  cost  of  1  A. 

"  26  A. 
"  1  R. 
"  20  sq.  rd. 
*'  5  sq.  rd, 

b-f-cH-dH-e  =  $4601.288  =    "     "  26  A.  1  R.  25  sq.  rd. 

Note.  —  In^  dividing,  in  examples  like  the  above,  it  is  unnecessary 
to  cany  out  the  work  further  than  to  mills. 

3.  How  much  will  13  lb.  4  oz.  10  dwt.  8  gr.  of  silver  cost 
at  $13.37  per  lb.  ? 

4.  How  much  will  9  T.  15  cwt.  1  qr.  14  lb.  of  English 
bar  i^:on  cost,  at  $83,625  per  ton,  reckoning  the  quarter  at 
28  lbs.  ?     (See  34,  c.) 

5.  How  much  will  it  cost  to  build  a  road  37  m.  5  fur.  30 
rd.  in  length,  at  the  rate  of  $2173.75  per  mile  ? 

WRITTEN  WORK. 

a  =  $  2173.75    =  cost  of  I  m. 


38  X  a  =  b  =  $82602.50  =  «  •  «  38  m. 
i  of  a  =  c  •-=  $  543.437  =  «  «  2  fur. 
^  of  c  =  d  =  $      67.929  =    "     «  10  rd. 

b  —  c  —  d  =  $81991.134  =  cost  of  37  m.  5  fur.  30  rd. 

6.  How  much  will  22  hhd.  21  gal.  3  qt.  1  pt.  of  wine  coat, 
at  $47,875  per  hhd.,  if  each  hogshead  contains  63  gallons  ? 

7.  How  much  will  24  bu.  1  pk.  6  qt.  of  wheat  cost,  at 
$1,375  per  bushel? 

8.  How  much  wiU  37  bu.  3  pk.  4  qt.  of  wheat  cost  at 
$1,625  per  bushel  ? 

9.  How  much  will  14  cwt.  2  qr.  15  lb.  of  coffee  cost,  at 
$12,625  per  cwt.  ? 

10.  A  man  bought  a  farm,  containing  137  A.  3  R.  30  sq. 
rd.,  at  $46.94  per  acre  ?     How  much  did  it  cost  him  ? 

11.  A  trader  bought  184  bu.  2  pk.  6  qt.  of  Indian  com,  at 
$-875  per  bushel.     How  much  did  it  cost  him  ? 


22G  PRACTICE. 

12.  How  many  rods  will  a  man  travel  in  13  h.  20  m.  45 
sec,  if  he  travels  1289  rods  per  hour  ? 

13.  What  will  24  A.  1  R.  25  sq.  rod  cost,  at  $47.98  per 
acre? 

14.  What  will  17  T.  15  cwt.  2  qr.  7  lb.  of  railroad  iron 
cost,  at  $47.38  per  ton  ? 

15.  How  much  will  4  lb.  11  oz.  19  dwt.  12  gr.  of  silver 
cost,  at  $13.96  per  pound  ? 

16.  What  will  428  bu.  of  oats  cost,  at  $.49  per  bushel? 

Suggestion.  —  At  49  cents  per  bushel,  428  bushels  will  cost  428  half 
dollars,  minus  428  cents. 

17.  How  much  will  43  volumes  cost,  at  $1.99  per  volume? 

18.  How  much  will  48  bbls.  of  flour  cost,  at  $8.75  per 
barrel  ? 

Suggestion.  —  At  $8.75  per  barrel,  48  barrels  will  cost  48  times  $9  — 
48  times  ^  of  a  dollar. 

19.  How  much  will  32  yds.  of  cloth  cost,  at  $1,875  per 
yard? 

Suggestion.  —  At  $1,875  per  yard,  32  yards  will  cost  32  times  $2  — 

32  times  ^  of  a  dollar. 

» 

20.  How  much  will  747  lbs.  of  tea  cost,  at  $.66f  per  lb.  ? 

21.  How  much  will  8  cords  of  wood  cost,  at  $4.94  per 
cord  ? 

22.  How  much  will  4  acres  of  land  cost,  at  $98.75  pei 
acre? 

23.  What  will  1728  yd.  of  cloth  cost,  at  1  s,  8  d.  per  yard? 

Suggestion.     1  s.S  i.  =  i^  of  £1. 

24.  What  will  857  yd.  of  broadcloth  cost,  at  16  s.  8  d.  per 
yard? 

Suggestion.    16  s.  8  d.  =  £1  --  3  s.  4  d.  =  £1  —  ^  of  £1. 

Or,  16  8.  8  d.  =  10  8.  +  6  s.  8  d.  =  i^  of  £1  4-  ^  of  £1. 

25.  What  will  1478  reams  of  paper  cost,  at  13  s.  4  d.  per 
ream? 

Suggfution.     13  s.  4  d.  =  £1  —  6  8.  8  d.  =  £1  —  \  oi  £\. 


FRACTICE.  2W 

26.  What  will  737  yd.  of  black  silk  cost,  at  17  s.  6  d.  per 
yard? 

Suggestion.    2  s.  6  d.  =  i  of  £1. 

27.  What  will  138  yd.  of  silk  velvet  cost,  at  £1  5  s.  6  d. 
per  yard  ? 

WRITTEN  WOBK. 

a  =  £138  =  cost  at  £1  per  yd. 

^  of  a  =  b  =  X  34  10  s.  =    "    "5  s.  per  yd. 

lV  of  b  =  c  =  £    3     9  s.  =    "    "  6  d.  per  yd. 

a  -f-  b  +  c  =  £175  19  s.  =  cost  at  £1  5  s.  6  d.  per  yard. 

28.  What  will  13  coats  cost,  at  £3  16  s.  8  d.  each  ? 
\ 

WRITTEN    WORK. 

a  =  £13  s=  cost  at  £1  each. 

4  -f-  a  =  b  =  £52  »==     «     «c  £4  g^ch. 

1^  of  a  =  c  =  £  8  13  s.  4  d.  =  cost  at  3  s.  4  d.  each, 

b  —  c  =  £43  6  8.  8  d. 

29.  What  will  24  tons  of  iron  cost,  at  £8  2  s.  6  d.  per  ton  ? 

30.  What  will  65  pieces  of  broadcloth  cost,  at  £29  19  s. 
8  d.  per  piece  ? 

31.  What  will  45  pieces  of  Irish  linen  cost,  at  £3  1  s.  4  d. 
per  piece  ? 

32.  What  will  96  tons  of  iron  cost,  at  £1  11  s.  4  d.  per 
ton? 

33.  If  14  yd.  1  qr.  2  na.  1  in.  of  cloth  cost  £12  13  s.  8  d., 
what  will  28  yd.  3  qr.  2  in.  of  cloth  cost  ? 

Suggestion.    28  yd.  3  qr.  2  in.  =  twice  14  yd.  1  qr.  2  na.  1  in. 

34.  If  7  lb.  3  oz.  4  dr.  of  sugar  cost  2  s.  9  d.  1  qr.,  what 
will  211b.  9oz.  12  dr.  cost? 

35.  If  17  bu.  1  pk.  2  qt.  of  corn  cost  $9.39,  how  much  will 
61  bu.  3pk.  6qt.  cost? 

36.  If  13  gal.  3  qt.  1  pt.  1  gi.  of  molasses  cost  $3.87,  how 
much  will  41  gal.  2  qt.  1  pt.  3  gi.  cost  ? 

37.  If  17  A.  3  R.  7  rd.  of  land  cost  $849.29,  what  will 
88  A.  3  B.  35  rd.  cost  ? 

19* 


2*22  RATIO    AND    PROPORTION. 

38.  If  98  yd.  3  qr.  2  na.  of  cloth  cost  $44.38,  what  will 
49  yd.  1  qr.  3  na.  cost  ? 

39.  If  68  bu.  1  pk.  6  qt.  of  wheat  cost  $114.48,  what  will 
22  bu.  3  pk.  2  qt.  cost  ? 

40.  if  23  lb.   13  oz.  8  dl-.  of  potash  can  be  bought  for 
$2.12,  how  much  can  be  bought  for  $16.96  ? 

41.  If  9  T.  14  cwt.  2  qr.  17  lb.  of  hay  can  be  bought  for 
$171.32,  how  much  can  be  bought  for  $513.96  ? 


SECTION    XII. 

RATIO   AND   PROPORTION. 

Note.  —  If  the  teacher  prefers  it,  the  pupil  may  omit  this  section 
and  the  two  following  it,  till  after  he  has  mastered  the  sections  on  inters 
est  and  the  subjects  pertaining  to  business  life. 

1^0*    Definitions  and  Illustrations  of  Ratio. 

(a.)  Ratio  is  the  part  which  one  number  is  of  another,  or 
the  quotient  of  one  number  divided  by  another :  — 

Thus,  the  ratio  of  5  to  6  is  |^,  or  1-^,  because  6  equals  |-  of  5,  equaU 
l\  times  5  ;  or  because  6  -i-  5  =  |-  =  13^. 

The  ratio  of  3  to  12  is  V'>  or  4,  because  12  equals  -^^  of  3  =  4  times 
3  ;  or  because  12  -^  3  =  -^  =  4 

Note.  —  Some  writers  consider  that  the  ratio  of  5  to  6,  or  of  3  tc 
12,  is  the  part  which  5  is  of  6,  or  3  is  of  12,  instead  of  the  part  which  6 
is  of  5,  or  12  is  of  3,  as  given  above. 

The  difference  is  not  practically  of  so  much  consequence  as  would  at 
first  appear  to  be  the  case,  for  the  term  ratio  is  almost  invariably  used 
in  some  such  connection  as  the  following :  "  The  ratio  of  4  to  6  equals 
the  ratio  of  10  to  15,"  where,  by  the  first  interpretation,  we  have  f  =  -f^, 
or  ^,  =  f ;  and  by  the  second,  ^  =  \^,  or  §,  =  f,  both  of  whiih  are 
manifestly  true.  It  should,  then,  be  borne  in  mind,  that  while  the  first 
interpretation  is  the  one  usually  adopted,  the  second  may  be  substituted 
for  it,  in  any  case  where  tke  change  may  seem  desirablo. 


i 


RATIO    AND    PROPORTION.  223 

{b.)  A  ratio  can  always  be  established  between  abstract 
numbers,  but  it  can  only  exist  between  conci-ete  numbers 
when  they  are  of  the  same  denomination  ;  for  one  quantity 
can  never  be  a  fractional  part  of  another  quantity  of  a  differ- 
ent kind. 

Thus,  8  apples  is  no  part  of  4  pears,  and  hence  has  no  ratio  to  it. 

(c.)    By  the  above  illustrations,  it  appears  that  every  ratio 

is  a  true  fraction,  and  may  be  written  and  dealt  with  as  such. 

(d.)    Ratios  are,  however,  usually  expressed  by  writing  one 

number  after  the  other,  and  placing  two  dots  between  them, 

thus :  — 

The  ratio  of  4  to  6  =  4  :  6,  or  f . 
The  ratio  of  9  to  7  =  9  :  7,  or  |-. 

(e.)  The  numbers  which  form  any  ratio  are  called  terms 
of  the  ratio ;  the  first  number,  or  term,  is  called  the  antece' 
dent  of  the  ratio  ;  and  the  second  number,  or  term,  is  called 
the  consequent  of  the  ratio. 

Thus,  in  the  ratio  9  :  7,  9  is  the  consequent,  and  7  the  antecedent. 

(/.)    Ratios,  like  fractions,  may  be  simple,  complex,  or 

COMPOUND. 

{g.)  A  SIMPLE  RATIO  is  the  ratio  of  two  entire  numbers  ; 
as,  5  :  7,  4  :  9,  or  I,  |. 

(A.)    A  COMPLEX   RATIO  is  the   ratio   of  two  fractional 

numbers ;  as,  |  to  2^-,  ^  :  3|,  or  M,  ?i. 

(i.)  A  COMPOUND  RATIO  is  the  indicated  product  of  two 
or  more  ratios ;  as  (5  :  7)  X  (8  :  3),  or  i  of  f ,  or  |  X  f- 

(J.)    A  compound  ratio  is  usually  expressed  by  writing  the 

ratios  which  compose  it  under  each  other  :  — 

4  :  71 
Thus,  9  .  g  r  expresses  that  the  product  of  the  two  ratios,  4  :  7  and 

9  ;  8,  is  to  be  obtained ;  or,  which  is  the  same  thing,  it  means  J  of  |-, 
orjxi 

1.57,    Rediiction  of  Ratios, 

(a.)  Since  ratios  are  really  fractions,  the  principles  in- 
yolved  in  all  operations  upon  ihem  are  precisely  the  same  as 


224 


RATIO    AND    PROPORTION. 


those  involved  in  the  corresponding  operations  on  other  frac- 
tions. 

(b.)  Hence,  multiplying  or  dividing  both  terms  of  a  ratio 
by  the  same  number  will  not  alter  its  value. 

1.    Reduce  4  :  6  to  its  lowest  terms. 

Solution.  —  Dividing  both  terms  by  2  gives  4:6  =  2:3.  ThiB 
may  be  proved  thus  :4:6  =  |-  =  ^  =  2:3. 

Reduce  each  of  the  following  ratios  to  its  lowest  terms  :  -^ 


2.  6:9. 

3.  12  :  15. 

4.  21  :  35. 

10. 
11. 
12. 

13. 


24  :  48. 
16  :  12. 

8  :  6. 


18  :  15. 
36  :  54. 


16  X  9  :  8  X  12. 
15  X  3  X  7  :  35  X  9  X  4. 
14  X  8  X  5  :  49  X  10  X  12. 
20  X  7  X  45  ;  9  X  35  X  10. 


14.    Reduce  4|  :  Q^jj  to  a  simple  ratio. 

First  Solution.  —  Multiplying  both  terms  by  10,  the  least  common 
multiple  of  5  and  10,  (see  149,  e.)  gives  42  :  63  =  2  :  3. 

2 


Second  Solution. 

3 


H  :  6^  = 


21  ,  63 
5    •   10 


=  ^t  X  X0  :  $ 


X  0;g(  =  2    :    3.      (See  149,  c.) 
Reduce  the  following  complex  ratios  to  simple  ones  : 


15. 
16. 
17. 
18. 
19. 


H  :  4|. 
.03  :  5.7. 

6f 
005. 


Ait 
.02 


20. 
21. 
22. 
23. 


3^  :  n. 
If  :  1,2^. 
2.0007  :  2000.7 
.0025  :  2500. 


5  :     8 

24.   Reduce  12  :  14 

21  :  25 


to  a  simple  ratio. 


WRITTEN    WOKK. 

3  3       2       2        5 

X  X^  X  ^t  :  $xUx^$  =  9  :  20. 


RATIO    AND    PROPORTION. 


225 


Proof. 


9  :  20. 


8     .  14     ,  25 
5  ^^  r2  ^^  21 


2    2     5 

20 

$   X   X^   X   ^t  ' 
3     3 

^  T 

Reduce  each  of  the  following  compound  ratios  to  simple 


ones : — 

'  8  :  12 

25. 

<  9  :  10 

15:6 

"4:7 

26. 

^  6  :  25 

.15  :  24 

' 24  :  49 

27. 

^  25  :  60 

1 21  :  30 

28. 


29. 


30. 


11 

45 
63 

'i 

8 

H 
H 


Note.  —  Erom  the  above,  it  is  obvious  that  a  compound  ratio  may 
be  reduced  to  a  simple  one,  by  making  all  its  antecedents  factors  of  the 
new  antecedent,  and  all  its  consequents  factors  of  the  new  consequent, 
and  then  cancelling  all  the  factors  common  to  both  terms. 

1^8,    Definitions  and  Illustrations  of  Proportion, 

(a.)    A  PROPORTION  is  an  equality  of  ratios. 

{h.)  Proportions  may  be  simple,  complex,  or  com- 
pound. 

(c.)  A  SIMPLE  PROPORTION  expresses  the  equality  of  two 
simple  ratios. 

(d.)  A  COMPLEX  PROPORTION  expresscs  the  equality  of 
two  complex  ratios,  or  of  a  complex  and  simple  ratio. 

(e.)  A  COMPOUND  PROPORTION  cxpresscs  the  equality  of 
two  compound  ratios,  or  of  a  compound  and  simple  ratio. 

{f.)  A  proportion  is  expressed  by  writing  two  ratios  one 
after  the  other,  and  placing  four  dots  between  them. 

Thus,  4  :  6  :  :  12  :  18  is  a  proportion  which  expresses  that  the  ratio 
of  4  to  6  equals  the  ratio  of  12  to  18.  It  would  be  read,  4  is  /o  6  as  13 
is  to  18,  or  6  is  the  same  part  of  4  that  18  is  of  12, 

3^  :  5^  :  :  2^^  :  3^  expresses  that  the  ratio  of  3^  to  5^  equals  the 


226  RATIO    AND    PROPORTION. 

ratio  of  2-i^o  to  S-^.    It  would  be  road,  3^  is  to  b^  as  2-iV  is  to  3-J,  or  6^ 
is  the  same  pari  of  34-  that  3-^  is  of  2Yij. 
The  compound  proportion  4  :     7  ^ 

9  :  1 6  >  :  :  3  :  2,  expresses  that  the  ratio 
14  :    sJ 
of4X9X14to7X16X3  equals  the  ratio  of  3  :  2.    It  would  be 
read,  4  times  9  times  14  is  to  7  times  16  times  S  as  3  is  to  2,  or  7  times  16 
times  3  is  the  same  part  of  4  times  9  times  14  that  2  is  of  3. 

'(t^  )  The  sign  of  equality  is  sometimes  used  instead  of  the 
four  dots. 

Thus,  instead  of  4  :  6  :  :  2  :  3,  we  may  have  4:6  =  2:3. 

(h.)  Every  proportion  may  be  expressed  as  the  equality 
of  two  fractions. 

Thus,  we  may  express  the  first  of  the  above  by  |-  =  -ff-,  the  second 
^y  ^  =  f^'  ^°^  t^«  t^i^d  by  {  of  -^  of  t\  =  f . 

(i.)  The  outside  terms  (i.  e.,  the  first  and  fourth)  of  a  pro- 
portion are  called  the  extremes,  and  the  inside  terms  (i.  e., 
the  second  and  third)  the  means  of  the  proportion. 

Thus,  in  the  first  proportion  above  given,  4  and  18  are  the  extremes. 
6  and  12  are  the  means:  in  the  second,  3^  and  3-5- are  the  extremes 
5g-  and  2x\y  are  the  means;  in  the  third,  4  X  9  X  14  and  2  are  the 
extremes,  and  7X16X3  and  3  are  the  means. 


1^9.    Method  of  finding  a  missing  Term. 

(a.)  If  any  term  of  a  proportion  is  omitted,  it  may  easily 
be  supplied ;  for  from  the  nature  of  a  proportion,  it  follows 
that,  — 

First.  The  missing  antecedent  of  any  proportion  must  he 
the  same  part  of  its  consequent  that  the  given  antecedent  is  of 
its  consequent. 

vSecond.  The  missing  consequent  of  any  proportion  must  be 
(he  same  part  of  its  antecedent  thai  the  given  conseqv£nt  is  of 
its  antecedent. 

1.    Find  the  missing  term  of  the  proportion  5  :  7  : :  25  :  — 


RATIO    AND    PROPORTION.  227 

Solution.  —  The  missing  tenn  is  the  same  part  of  25  that  7  is  of  5  ; 

7  X  25 
i.  e^  it  is  T  of  25,  which  is  35.*  Hence,  5  :  7  :  :  25  :  — - —  =.35. 

5 

2.  Find  the  missing  term  of  the  proportion  9:14::  — 
:  49. 

Solution. —  The  missing  term  is  the  same  part  of  49  that  9  is  of  14  ; 

9  X  49 

i.  e.,  it  is  ^4  of  49,  which  is  31^.     Hence,  9:14::  =  31^  :  49. 

14 

3.  Find  the  missing  term    of  the  proportion  4§  :  3^  : : 

n  ■•  — 

Solution.  —  The  missing  term  is  the  same  part  of  7^  that  3^  is  of 
4§ ;  i.  e.,  it  is  — 

9 

Hence,4f  :3i::7i:?ii^  =  5f. 

4.  Find  the  missing  term  of  the  proportion 

7  :     8  V  :  :  25  :  — .  J|. 

15  :  14  J 

Solution.  —  The  missing  term  is  tlie  same  part  of  25  that  9  X  8  X 
14  is  of  4  X  7  X  15;  i.  e.,  it  is 

3       2        2  5 

9  X  8  X  14   ,  ^_,      0  X  ^  X  a:^  X  ^^      ^^ 

-; = =-=  01  25  = 7 Z2 -To. —  =  oU. 

$ 

Hence,    4  :     9 1  9  X  8  X  U  X  25       ^^ 

^'-     n"-^^'        4X7X15       =^"- 
15  :  Uj 

Find  the  missing  term  of  each  of  the  following  propor- 
tions :  — 


5.  9  :  12  :  :  6  :  — . 

6.  4^  :  10^  :  :  26  :  — . 

7.  36  :  48  : :  —  :  60. 

8.  53  :  42  :  :  —  :  3f 


9.  4  :  —  :  :  20  :  28. 

10.  9|  :  —  :  :  7  :  28. 

11.  —  :  6  ::  13  :  52. 

12.  —  :  .9   ::   4.9   :   .063 


228  RATIO    AND    PROPORTION. 


13.  <^  8  :  12  ^  : :  54  :  — .      15.  ^  42   :  32  S  : :  —  :  64 


3  :     5^ 

8  :  12  y 

:  :  54  :  — . 

9  :     7) 

H'-n  ] 

5^:2/^1 

^::144:^. 

4    :6      J 

21   : 

17^ 

42   : 

32  S  :: 

51   : 

54  J 

.03: 

.007^1 

3.5  : 

500  >  : 

.9  : 

2.7  J 

14    ^  54.:2^V  ^::144:^.     16.  ^  3.5  :  500  ^  : :  —  :  30. 


160.    Relations  of  the  Terms. 

(a.)  From  the  foregoing  explanations  and  illustrations,  we 
\\  ij  infer  that,  — 

first.  Either  extreme  is  equal  to  the  quotient  obtained  hy 
dividing  the  product  of  the  means  hy  the  other  extreme. 

Second.  Either  mean  is  equal  to  the  quotient  obtained  hy 
dividing  the  product  of  the  extremes  by  the  other  mean. 

(h.)    Hence,  in  a  proportion  — 

The  product  of  the  means  is  equal  to  the  product  of  tlie 
extremes. 

161.    Practical  Problems. 

(a.)  The  forming  of  a  proportion  from  the  conditions  of  a 
problem  is  called  stating  it. 

{b.)  In  stating  a  proportion,  it  is  customary  to  write  the 
number  which  is  of  the  same  denomination  as  the  answer  for 
the  third  term. 

1.   If  8  books  cost  $4.32,  what  will  11  books  cost? 

Solution.  —  Since  the  answer  is  to  be  in  dollars,  we  write  $4.32  a^ 
the  third  term,  and  this  being  the  price  of  8  books,  11  books  will  cost 
the  same  part  of  this  that  11  is  of  8,  and  we  therefore  make  1 1  the 
second  term,  and  8  the  first    Hence,  the  following  statement :  — 

4  39  V    11 

8  :  11  : :  $4.32  :  $I±'_^-_  =  $5.94  =  Ans. 
8 

Note.  —  The  above  solution  is  really  equivalent  to  the  following: 
Since  the  answer  is  to  be  in  dollars,  we  write  $4.32,  as  the  number  on 
which  the  operation  is  to  be  performed.  If  8  books  cost  so  much,  1 1 
books  will  cost  -V"  of  tbls,  which  may  be  found  by  making  1 1  a  factor 


RATIO    AND    PROPORTION.  229 

of  the  numerator,  and  8  a  factor  of  the  denominator.    Hence,  the  fol- 
lowing written  work :  — 

^4.32  X  11 


8 


=  $5.94. 


2.  If  it  takes  18  men  7  days  to  pei-form  a  piece  of  work, 
how  many  men  will  it  take  to  perform  it  in  9  days  ? 

Solution.  —  Since  the  answer  is  to  represent  the  number  of  men,  we 

write  18  men  as  the  third  term.    But  since  it  will  take  7  times  as  many 

men  to  do  it  in  1  day  as  it  will  to  do  it  in  7  days,  and  ^  as  many  to  do  it 

in  9  days  as  it  will  to  do  it  in  1  day,  it  follows  that  the  answer  is  the 

same  part  of  18  that  7  is  of  9.     Hence, 

7X18 

9  :  7  :  :  18  :  men  =  14  men  =  Ans. 

9 

Note.  —  "When  the  pupil  is  familiar  with  the  full  form,  he  may 
abbreviate,  thus  :  Since  the  answer  is  to  represent  a  number  of  men, 
we  write  18  men  as  the  third  term.  But  as  this  is  the  number  of  men 
which  it  takes  to  do  it  in  7  days,  it  will  take  the  same  part  of  this  num- 
ber to  do  it  in  9  days  that  7  is  of  9.  This  gives  the  same  proportion  as 
before. 

3.  If  it  takes  a  man  7^  days,  of  10^  hours  each,  to  earn 
$26^,  how  many  days,  of  9^  hours  each,"will  it  take  him  to 
earn  the  same  sum  ? 

Solution.  —  Since  the  answer  is  to  be  in  days,  we  write  7^  for  the 
third  term.  If,  when  the  days  are  10^  hours  long,  it  takes  so  many 
days,  when  they  are  9-^  hours  long  it  will  take  the  same  part  of  this 
number  of  days  that  10^  is  of  9-5-,  which  will  give  9-^  for  the  first  term, 
and  10^  for  the  second.    Hence,  9-^  :  10^  : :  72-  :  — . 

(c.)  After  having  written  the  third  term,  we  can  tell  in 
what  order  to  arrange  the  two  remaining  numbers,  by  observ- 
ing that  when  the  conditions  of  the  question  are  such  as  to 
require  an  answer  greater  than  the  third  term,  the  larger 
number  will  be  the  second  term,  and  the  smaller  the  first ; 
and  that  when  they  are  such  as  to  require  an  answer  less 
than  the  third  term,  the  smaller  number  will  be  the  second 
term,  and  the  larger  the  first. 

Thus,  in  the  first  example,  having  written  $4.32  as  the  third  term, 
we  may  observe  that  the  cost  of  11  books  will  be  greater  than  the  cost 
of  8  books,  and  that  the  ratio  is  8  :  11. 
20 


230  RATIO   AND   PROPORTION. 

In  the  second  example,  having  written  18  for  the  third  term,  we  may 
observe  that  it  will  take  a  less  number  of  men  to  perform  the  work  ia  9 
days  than  to  perform  it  in  7  days,  and  that  the  ratio  is  9  :  7. 

In  the  third  example,  having  written  7^  days  as  the  third  term,  we 
may  observe  that  it  will  take  more  days  to  earn  the  money  when  they 
are  9-5^  hours  long,  than  when  they  are  10^  hours  long,  and  that  the 
ratio  is  9-5-  :  10^. 

4.  If  24  cows  can  be  bought  for  $486,  for  how  much  can 
17  cows  be  bought  ? 

o.  If  17  cows  can  be  bought  for  $344.25,  for  how  much 
can  24  cows  be  bought  ? 

6.  If  24  cows  can  be  bought  for  $486,  how  many  can  be 
bought  for  $344.25  ? 

7.  If  17  cows  can  be  bought  for  $344.25,  how  many  can 
be  bought  for  $486  ? 

8.  How  much  would  it  cost  to  transport  9  cwt.  2  qrs.  13  lb. 
of  merchandise  from  Providence  to  Boston,  if  it  costs  $6.43 
to  transport  12  cwt.  3  qrs.  11  lb.  the  same  distance? 

9.  How  many  men  will  it  take  to  perform  a  piece  of  work 
in  6f  days,  which  it  will  take  42  men  1 2f  days  to  perform  ? 

10.  How  many  days  will  it  take  81  men  to  perform  a 
piece  of  work  which  42  men  can  do  in  12f  days  ? 

11.  How  many  men  will  it  take  to  do  a  piece  of  work  in 
12f  days  which  81  men  can  do  in  6§  days  ? 

12.  How  many  days  will  it  take  42  men  to  do  a  piece  of 
work  which  81  men  can  do  in  6f  days  ? 

13.  John  walks  3f  miles  per  hour,  and  William  walks  3^ 
miles  per  hour.  How  many  hours  will  it  take  John  to  walk 
as  far  as  WilUam  can  walk  in  8  hours  ? 

14.  If  9^  yards  of  cloth  can  be  bought  for  $44^,  how 
many  yards  can  be  bought  for  $33^  ? 


lOd.    Problems  in  Compound  Proportion. 

1.   If  6  boxes  of  soap,  each  holding  9  pounds,  cost  $4.59 
how  much  will  11  boxes,  each  holding  12  pounds,  cost? 


RATIO    AND    PROPORTION.  231 

Solution,  —  Observing  that  the  answer  is  to  be  in  dollars^  we  writo 
$4.59  as  the  third  term.  As  this  is  the  cost  of  6  boxes  of  soap,  1 1 
boxes  will  cost  the  same  part  of  this  that  11  is  of  6,  which  may  be  ex- 
pressed by  making  11  the  second  term,  and  6  the  first.  But  if  boxes 
each  holding  9  pounds  cost  so  much,  boxes  each  holding  12  pounds 
will  cost  the  same  part  of  this  that  12  is  of  9,  which  may  be  expressed 
by  making  12  a  factor  of  the  second  term,  and  9  a  factor  of  the  first. 
Heace  the  following  compound  proportion  :  — 

51  £ 

9  :   12}  •  •   *-^^    •    $r^^ -   ^^-^^ 

2.  IF  6  boxes  of  soap,  each  holding  9  pounds,  can  be 
bought  for  $4.59,  how  many  boxes,  each  holding  12  pounds, 
can  be  bought  for  $11.22. 

Solution.  —  We  first  write  6  boxes  for  the  third  term.  But  as  $11.22, 
or  1122  cents,  will  buy  the  same  part  of  what  $4.59,  or  459  cents,  will 
buy,  that  1122  is  of  459,  we  make  1122  the  second  term,  and  459  the 
first.  But  these  boxes  hold  9  pounds  each,  and  of  boxes  holding  12 
pounds  each,  only  -^^  as  many  can  be  bought,  which  may  be  expressed 
by  making  9  a  factor  of  the  second  term,  and  12  a  factor  of  the  first. 
Hence  the  proportion 

11 
^^ 

459  :  ll^^l  0  X  Xtm  X  0  _   ,, 

12  :  9        I  •  •  ^  •       ^00  X  1^        ~ 

(«.)  It  will  be  seen  by  the  above,  that,  after  writing  the 
third  term,  we  consider  what  ratio  an  answer  depending  on 
any  two  similar  conditions  of  the  question  would  bear  to  that 
third  term,  and  that,  after  writing  this  ratio,  we  consider  what 
ratio  an  answer  depending  upon  any  other  two  conditions 
similar  to  each  other  would  bear  to  this,  and  so  on  till  all  the 
conditions  are  considered.  Then,  by  solving  the  resulting 
compound  proportion,  the  answer  may  be  easily  obtained. 

(b.)  If  in  any  case  the  pupil  is  in  doubt  how  to  arrange 
the  terms  of  a  ratio,  he  may  determine  it  by  the  method  indi 
cated  in  161,  c. 


282  RATIO    AND    PROPORTION. 

Examples  like  those  last  explained  are  really  equivalent  to 
two  or  more  examples  in  simple  proportion. 

Thus,  the  first  is  equivalent  to  the  two  following :  — 

1.  If  6  boxes  of  soap  cost  $-4.59,  how  much  will  11  boxes  cost  ? 

2.  If  a  certain  number  of  boxes  of  soap,  each  containing  9  pounds, 
cost  the  answer  to  the  last  question,  how  much  will  the  same  number  of 
boxes,  each  containing  12  pounds,  cost  1     (See  note  on  215th  page.) 

3.  If  it  costs  $36  to  transport  14  tons  of  merchandise  54 
miles,  how  much  will  it  cost  to  transport  18  tons  49  miles  ? 

4.  K  14  tons  of  merchandise  can  be  transported  54  miles 
for  $36,  how  many  tons  can  be  transported  49  miles  for  $42  ? 

5.  If  14  tons  of  merchandise  can  be  transported  54  miles 
for  $36,  how  many  miles  can  18  tons  be  transported  for  $42? 

6.  If  18  tons  of  merchandise  can  be  transported  49  miles 
for  $42,  how  many  tons  can  be  transported  54  miles  for  $36  ? 

7.  If  18  tons  of  merchandise  can  be  transported  49  miles 
for  $42,  how  many  miles  can  14  tons  be  transported  for  $36  ? 

8.  If  it  costs  $42  to  transport  18  tons  of  merchandise  49 
miles,  how  much  will  it  cost  to  transport  14  tons  54  miles  ? 

9.  If  8  men  can  earn  $216  in  3  weeks,  how  many  dollars 
can  12  men  earn  in  2  weeks  ? 

10.  If  it  takes  24  pounds  of  cotton  to  make  2  pieces  of 
sheeting,  each  containing  33  yards,  1^  yard  wide,  how  many 
pounds  of  cotton  will  it  take  to  make  11  pieces  of  sheeting, 
each  containing  27  yards,  1^  yard  wide  ? 

11.  If  it  takes  45  men  8  weeks,  working  5^  days  per 
week,  and  10  hours  per  day,  to  build  a  road  3^  miles  long 
and  4  rods  wide,  how  many  weeks  will  it  take  63  men,  work- 
ing 4^  days  per  week,  and  1 1  hours  per  day,  to  build  a  road 
12§  miles  long  and  3  rods  wide  ? 

12.  A  company  of  40  men  agree  to  perform  a  piece  of 
work  in  50  days,  but  after  working  9  hours  per  day  for  30 
days,  they  find  that  they  have  done  but  half  tlie  work.  How 
many  more  men  must  they  employ,  that,  by  working  10  houn 
per  day,  they  may  finish  the  job  according  to  agreement  ? 

13.  If  9  men,  by  working  8  hours  per  day,  can  mow  3l 


DUODECIMAL    FRACTIONS.  233 

acres  of  grass  in  2^  days,  how  many  acres  of  grass  can  5 
men  mow  in  3|  days,  by  working  7^  hours  per  day  ? 

14.  How  many  bottles  can  be  filled  from  12  casks  of  wine, 
each  cask  containing  126  gallons,  if  1344:  bottles  can  be  filled 
from  3  casks,  each  cask  containing  84  gallons  ? 

15.  If,  when  land  is  worth  $46f  per  acre,  a  lot  of  land,  35 
rods  long  and  24  rods  wide,  is  given  for  8  piles  of  wood,  each 
45  feet  long,  5  feet  high,  and  4  feet  wide,  how  much  ought 
land  to  be  worth  an  acre,  when  3  lots,  each  32  rods  long,  and 
25  rods  wide,  are  given  for  21  piles  of  wood,  each  37^  feet 
long,  4f  feet  high,  and  4  feet  wide  ? 


SECTION  XIII. 
163.   DUODECIMAL   FRACTIONS. 

m 
(«.)    A  DUODECIMAL  FRACTION,  Or  simply  a  DUODECIMAL, 

is  a  fraction  whose  denominator  is  12,  or  some  power  of  12. 

(6.)  The  denominator  of  a  duodecimal  is  not  written,  but 
is  indicated  by  one  or  more  marks,  or  accents,  placed  at  the 
right  of  the  numerator,  and  a  little  above  it 

Thus,4'  =  TV5   7"  =  Ti^;    11'"  =  tI^;   &c 

(c.)  In  reading  duodecimals,  twelfths  are  usually  called 
PRIMES  ;  one-hundred-and-forty-fourths  are  called  seconds  ; 
&c. 

Thus,  4'  is  read, ^«r  primes;  7"  is  read,  seven  seconds ;  11'"  is  read, 
eleven  thirds  ;  &c. 

{d.)  Duodecimals  are  employed  in  measuring  lengths,  sur- 
faces, and  solids ;  so  that  the  unit  of  measure  is  a  foot  of 
either  long,  square,  or  cubic  measure,  according  to  the  nature 
of  the  thing  measured. 

(e.)  Since  a  linear  inch  equals  j^  of  a  linear  foot,  a  square 
inch  Y^^-  of  a  square  foot,  and  a  cubic  inch  ytVf  ^^  ^  cubic 
20* 


2:>4  DUODECIMAL    FRACTIONS.  ^ 

foot,  it  follows  that  in  long  measure  the  inch  is  represented  by 
the  prime,  in  square  measure  by  the  second,  and  in  cubic 
measure  by  the  third. 

(/)  Duodecimals  may  be  added,  subtracted,  multiplied, 
and  divided  as  other  fractions  are.  In  performing  thesA 
operations,  it  is  necessary  to  notice  that  a  unit  of  any  denomi- 
nation equals  12  units  of  the  next  lower,  and  -^^  of  a  unit  of 
the  next  higher  denomination  ;  also,  that  1'  X  1'  =^  tV  X 
tV.  ==  tU,  or  1";  that  1"  X  1'  =  yii  X  tV  =  ttW,  or 
1'";    &c. 

163.    Problems. 

1.  What  are  the  contents  of  a  board  13  ft.  4'  long,  and  2 
ft.  b'  wide  ? 

Reasoning  Process.—  If  the  board  was  13  ft.  4'  long  and  1  ft.  wide,  it 
would  contain  13  sq.  ft.  4' ;  but  being  2  ft.  5'  wide,  it  must  contain  2-<^ 
times  13  sq.  ft.  4'.     This  gives  the  following  written  work  :  — 
ft.        / 

13      4  -, 

2       5 


5       6       8" 
26       8 


32  2'  8"  =  32  sq.  ft.  32  sq.  in. 
Explanation.  —  First,  multiplying  by  5'  (i.  e.,  by  -f^)  we  have  5' 
times  4'  =  20"  =  1'  8",  (i.  e.,  A  times  A  =  A\  =  tV  +  tI ¥•) 
Writing  the  8",  we  have  5'  times  13  ft.  =  65',  and  1'  added,  =  66'  = 
5  ft.  6",  (i.  e.,  tV  times  13  ft.  =  f|-  ft.,  &c.,)  which  we  write.  Thea 
multiplying  by  2,  we  have  2  times  4'  =  8' ;  2  times  13  ft.  =  26  ft.  Add- 
ing these  products  gives  32  ft.  2'  8"  for  a  result,  which,  being  in  square 
measure,  =  32  sq.  ft.  32  sq.  in. 

Note. —  The  student  should  observe  that  in  performing  the  work, 
the  multiplier  is  always  to  be  regarded  as  an  abstract  number.  (See 
74,  g.)  The  expression,  "  multiplying  the  length  by  the  breadth," 
means  simply  that  the  number  representing  the  length  is  to  be  multi- 
plied by  the  number  representing  the  breadth. 

2.  What  are  tlie  contents  of  a  board  14ft.  5'  long  and  1  ft. 
11' wide? 

3.  What  are  the  contents  of  a  blackboard  17  ft,  9'  long 
and  5  ft.  3'  wide  ? 


DUODECIMAL    FRACTIONS.  235 

4.  What  are  the  contents  of  a  pile  of  wood  47  f^.  6'  long, 
3  ft.  9'  wide,  and  5  ft.  4'  high  ? 

Note.  —  These  examples  can  very  frequently  be  performed  more 
easily  by  reducing  the  duodecimals  to  vulgar  fractions.  Thus,  in  the 
last  example,  47  ft.  6'  =  47^-  ft.,  3  ft.  9  =  sf  ft.,  and  5  ft.  4'  =  5^  ft., 
which,  in  conformity  with  41,  Note,  would  give  — 

5         2 

471  X  3f  X  5^  =  95  X  X$  X  t^  ^  g^^  ^^^  ^^  ^  ^^^^^ 
'^         *         -^         ^  X  ^  X  ;^ 

5.  How  many  cubic  feet  of  earth  would  be  removed  in 
digging  a  cellar  15  ft.  9'  long,  13  ft.  4'  wide,  and  9  ft.  8'  deep  ? 

6.  How  many  square  yards  of  carpeting  will  it  take  to 
cover  a  floor  16  ft.  6'  long  and  15  ft.  4'  wide  ? 

7.  How  many  square  feet  and  inches  in  the  four  walls  of 
a  room  23  fl.  5'  long,  18  ft.  9'  wide,  and  14  ft.  3  in.  high  ? 

8.  How  many  cords  and  cord  feet  of  wood  in  a  load  8  ft. 
long,  3  ft.  11'  wide,  and  5  ft.  7'  high  ? 

9.  How  many  yards  of  plastering  are  there  in  the  top  and 
walls  of  a  room  16  ft.  4'  long,  13  ft.  2'  wide,  and  lift.  6' 
high,  allowing  for  three  windows,  each  5  ft.  8'  high  and  3  ft. 
3'  wide  ;  two  doors,  each  6  ft.  10'  high  and  2  ft.  9'  wide,  and 
for  a  mop  board,  9  inches  wide,  around  the  room  ? 

10.  Mr.  Jackson  owned  a  garden,  137  ft.  long  and  112  ft. 
6'  wide,  around  which  he  laid  out  a  gravel  walk  4  ft.  8'  wide. 
How  many  square  feet  did  the  walk  contain  ? 

Note.  —  The  space  used  for  the  walk  is  supposed  to  be  taken  from 
the  contents  of  tlie  garden. 

Suggestion.  —  To  understand  this  problem, 
draw  a  figure  like  the  one  annexed.  Then  sup- 
pose the  walks  along  the  sides,  A  B  and  D  C, 
to  be  first  built.  Each  of  them  will  be  137  ft. 
lopg  and  4  ft.  8'  wide,  and  together  they  will  be 
equivalent  to  a  walk  2  X  137,  or  274  ft.  long 
and  4  ft.  8'  wide.  But  in  building  these  walks, 
it  is  obvious  that  we  have  shortened  each  walk 
to  be  built  across  the  ends  by  just  twice  the  width  of  the  walk,  i.  e.,  by 
twice  4  ft.  8',  which  is  9  ft.  4'.     This  taken  from   112  ft.  6  in.,  the  width 


236  DUODKCTMAI.    IKACTIONS. 

of  the  garden,  leaves  103  ft.  2',  which  is  the  length  of  the  walk  which 
remains  to  be  built  across  each  of  the  two  ends.  Twice  103  ft.  2'  =s 
206  ft.  4',  which,  added  to  274  ft.,  =  480  ft.  4',  =  the  whole  length  of 
the  walk  around  the  garden. 

11.  A  man  bought  a  garden  spot,  143  ft.  4'  long  and  124 
ft.  8'  wide,  and  after  leaving  a  space  for  a  flower  bed,  2  (It.  6' 
wide,  all  around  it,  he  laid  out  a  walk,  3  ft.  10'  wide,  within 

''  the  flower  bed,  and  extending  around  the  garden.    How  many- 
feet  did  the  walk  contain  ? 

12.  A  man  bought  a  lot  of  land,  97  ft.  4'  long  and  73  ft. 
3'  wide,  at  the  rate  of  6  cents  per  sq.  ft.  He  built  a  tight 
board  fence,  6  ft.  2'  high,  around  it,  for  which  he  paid  3  cents 
per  sq.  ft.    ^ow  much  did  the  land  and  fence  cost  him  ? 

13.  How  many  yards  of  carpeting,  1^  yd.  wide,  will  it 
take  to  cover  a  floor  19  ft.  9'  long  and  15  ft.  6'  wide  ? 

14.  How  many  bricks,  each  8  in.  long,  4  in.  wide,  and  2 
in.  thick,  will  it  take  to  build  a  wall  47  ft.  long,  6  ft.  6'  high, 
and  1  ft.  8'  thick  ? 

Suggestion.  —  Since  8  in.  =  §  ft.,  4  in.  =  -^  ft.,  and  2  in.  =  ij  ft., 
each  brick  contains  §  X  ^  X  ^  =  aV  solid  feet,  and  it  will  take  27 
bricks  for  each  solid  foot  in  the  wall. 

15.  How  many  bricks,  each  8  in.  long,  4  in.  wide,  and  2  in. 
thick,  will  it  take  to  build  the  four  walls  of  a  house,  30  ft.  6 
long  and  24  ft.  8'  wide,  the  walls  to  be  1  ft.  thick  and  22  ft. 
4'  high,  allowing  for  1  door,  8  ft.  high  and  3  ft.  10'  wide ;  for^ 
2  doors,  each  8  ft.  high  and  3  ft.  G'  wide;  for  12  windows, 
each  6  ft.  high  and  3  ft.  8'  wide;  and  for  IG  windows,  each 
5  ft.  8'  high  and  3  ft.  6'  wide  ? 

Note.  —  The  student  should  be  careful  to  notice  how  the  thickness 
of  the  wall  will  affect  his  calculations.  He  must  apply  a  principle 
similar  to  that  applied  in  the  note  to  the  10th  example. 

16.  A  prison  wall  is  built  of  Quincy  granite,  and  extends 
completely  round  the  prison  yard,  except  that  a  space  is  left 
in  one  of  its  sides  for  an  iron  gate  12  ft.  wide  and  12  ft.  high. 
The  outside  length  of  the  wall  is  54  ft.  4',  its  breadth  49  ft. 
6',  and  its  height  18  ft.  Its  sides  are  4  ft.  3'  thick,  and  its 
ends  3  ft.  9'.     How  many  cubic  feet  in  the  walj  ? 


ALLIGATION.  237 

SECTION   XIV. 
ALLIGATION. 

1G41.    Definitions  and  Explanations. 

(c/.)  Merchants  and  others  often  find  it  convenient  to  mix 
articles  of  different  kinds  together,  so  as  to  obtain  a  compound 
differing  in  value  from  any  of  its  ingredients.  The  various 
problems  connected  with  the  subject  are  called  Problems  in 
Alligation. 

(h.)  Questions  in  alligation  are  usually  divided  into  two 
classes,  viz. :  First,  Alligation  Medial,  in  which  the 
quantities  of  the  several  ingredients  and  their  prices  are 
given,  and  we  are  required  to  find  the  pnce  of  the  mixture 
per  pound,  per  gallon,  or  per  bushel. 

Second,  Alligation  Alternate,  in  which  the  prices  of 
the  various  ingredients  are  given,  and  we  are  required  to  find 
what  quantities  of  each  must  be  taken  to  make  a  mixture 
having  a  given  value  per  pound,  per  bushel,  or  per  gallon. 

16^.    Prohlems. 

1.  A  trader  mixed  together  6  lb.  of  coffee  worth  10  cents 
per  pound,  4  lb.  worth  8  cents  per  lb.,  and  7  lb.  worth  Itf 
cents  per  lb.     How  much  was  the  mixture  worth  per  lb*  ? 

Solution.    6  lb.  at  10  cents  are  worth    60  cents. 
4  "    «     8     "        "        "        32     " 
y  t(     ((    ig     ((         c(         a.       112     " 

Hence,  17  lb.  are  worth  $2.04,  and  1  lb,  is  worth  -T^  ot 

$2.04,  which  is  12  cents  :=  Ans. 

2.  A  silversmith  melted  together  9  oz.  of  silver,  14  carats 
fine;  6  oz.,  18  carats  fine;  12  oz.,  22  carats  fine;  and  23  oz., 
24  carats  fine.     What  was  the  fineness  of  the  mixture  ? 

3.  A  dry  goods  dealer  sold  25  yd.  of  sheeting  at  9  cents 
per  yd. ;  30  yd.  of  shirting  at  10  cents  per  yd. ;  11  yd.  of 


238  ALLIGATION. 

delaine  at  18  cents  per  yd.  ;  9  yd^  of  gingham  at  22  cent? 
per  yd. ;  and  25  yd.  of  linen  at  40  cents  per  yd.  What  wa3 
the  average  price  of  the  whole  per  yard  ? 

4.  A  merchant  has  sugars  at  6,  7,  10,  and  13  cents  per 
pound,  of  which  he  wishes  to  make  a  mixture  such  that,  by 
selling  it  for  9  cents  per  pound,  he  will  neither  gain  nor  lose. 
How  many  pounds  of  each  kind  must  he  take  ? 

Solution.  —  It  is  obvious  that  by  selling  the  mixture  for  9  cents  per 
pound,  he  will  gain  3  cents  on  each  pound  which  he  puts  in  of  the  first 
kind,  and  2  cents  on  each  pound  of  the  second  kind  ;  that  he  will  lose 
1  cent  on  each  pound  of  the  third  kind,  and  4  cents  on  each  pound  of 
the  fourth  ;  and  further,  that  to  make  the  mixture  worth  just  9  cents 
per  pound,  he  must  take  such  proportions  of  the  several  kinds  as  will 
make  his  gains  equal  his  losses.  Moreover,  he  may  take  as  many 
pounds  as  he  chooses  of  the  kinds  which  cost  less  than  9  cents,  provided 
he  takes  enough  of  the  others  to  counterbalance  the  gain  on  them. 

Suppose  that  he  takes  8  lb.  of  the  first  kind,  and  11  lb.  of  the  second. 
Then,  since  on  1  lb.  of  the  first  he  gains  3  cents,  on  8  lb.  he  will  gain  8 
times  3  cents,  or  24  cents  ;  and  since  on  1  lb.  of  the  second  he  gains  2 
cents,  on  11  lb.  he  will  gain  11  times  2  cents,  or  22  cents,  which,  added 
to  the  24  cents,  gives  46  cents  as  the  sum  of  his  gains.  He  must,  there- 
fore, take  enough  of  the  other  kinds  to  lose  46  cents.  Suppose  he  takes 
10  lb.  of  that  at  10  cents.  Then,  since  on  1  lb.  he  loses  1  cent,  on  10 
lb.  he  will  lose  10  times  1  cent,  or  10  cents,  and  he  must  take  enough 
of  that  at  13  cents  to  lose  36  cents.  Since  on  1  lb.  he  loses  4  cents,  he 
must  take  as  many  pounds  to  lose  36  cents  as  there  are  times  4  in  36 
which  are  9  times.  Hence,  he  may  take  8  lb.  at  6  cents,  11-  lb.  at  7 
cents,  10  lb.  at  10  cents,  and  9  lb.  at  13  cents. 

The  following  is  a  convenient  form  of  writing  the  work  :  — 


Mean 
price. 

Prices  of 
ingredients 

Gain  or  loss 
per  lb. 

Quantities  taken. 

Sum  of  gains 
or  losses. 

9 

'    6     .     . 

.     7     .    . 

.     +3 
.     +2 

8  lb.  g.  24  cts.' 
11  "    g.  22    " 

\  46  cts.  gain. 

10     .    . 
.13    .     . 

.    —  1 
.    —  4 

10  "    1.   10    "  ] 
9  "    1.  36    "  j 

46  cts.  loss. 

Proof.     8  lb.  at    6  cents  are  worth  $  .48. 

11  "    "     7     "       "       "      $  .77. 

10  "    "10     •'       "        "      $1.00. 

9  "    "  13     "       "        "      $1.17. 

Making  38  lb.,  worth     ....       $3.42. 


ALLIGATION.  239 

But  38  lb.,  at  9  cents  per  lb.,  would  brino;  just  $3.42,  which  shows 
that  the  merchant  would  get  the  same  sum  by  selling  the  mixture  at  9 
cents  per  lb.  that  he  would  by  selling  the  ingredients  separately,  at  their 
respective  prices. 

(c.)  There  is  no  limit^o  the  number  of  answers  which 
may  be  obtained  to  such  questions  as  the  above ;  for  however 
many  or  few  pounds  of  any  kind  we  take,  we  may  take 
enough  of  other  kinds  to  counterbalance  the  gain  or  loss.  In 
the  solution,  we  may  as  well  consider  first  the  number  of 
pounds  to  be  taken  of  the  kinds  which  cost  less  than  the 
mean  rate,  as  of  those  which  cost  more. 

(d.)  Annexed  is  a  part  of  the  written  work  of  two  other 
solutions  to  the  above  example.  Let  the  student  complete 
and  explain  it,  and  also  prove  the  correctness  of  his  results. 


Quantities  taken.    Sum  of  gains  duantities  taken.     Sum  of  gains 
and  losses.  and  losses, 

lb.  g.  $.39]  29  1b.  g.  $  .87] 

7  lb.  g.  $.14  [  ^-53  S^'""-  83  lb.  g.  $1.66  J  ^2-53  gain. 
5  1b.  1.   $.05]  lb.  I.  ] 

12  1b.  1.   $.48P-^^^«^^-  26  1b.  1.   $1.04  1°^«- 


(e.)  If  it  should  be  necessary  to  make  a  mixture  contain- 
ing a  given  number  of  pounds,  we  should  first  get  an  answer 
to  the  question,  as  though  no  limit  had  been  specified,  and 
then  find  how  many  times  as  much  should  be  taken  to  give 
the  required  quantity. 

Suppose,  for  instance,  that  the  above  question  had  read,  "  How  many 
pounds  of  each  kind  must  he  take  to  make  a  mixture  of  100  lb.  worth 

9  cents  per  lb.,"  we  should  have  the  following  additional  work  :  — 

The  first  solution  gives  a  mixture  containing  38  lb.;  and  since  100  lb. 
=  2^g-  times  38  lb.,  we  must  take  2xf-  times  as  much  of  each  of  the 
former  ingredients  as  before,  which  would  give  2^'^  times  8  lb.,  or  21-yV 
lb.  of  the  first ;  2j^  times  1 1  lb.,  or  28^1  lb.  of  the  second  ;  2x1  times 

10  lb.,  or  26T^g-  lb.  of  the  third ;  and  2^1  times  9  lb.,  or  23^1  lb.  of  the 
fourth.  The  proof  is  the  same  as  though  these  quantities  had  been 
originally  selected. 

5.  A  trader  has  coffees  at  8,  10,  13,  and  15  cents  per  lb. 
How  many  pounds  of  each  may  he  take  to  make  a  mixture 
worth  12  cents  per  lb.  ? 


240  INTEREST. 

6.  A  trader  has  molasses  at  22,  25,  29,  and  33  cents  per 
gallon.  How  many  gallons  of  each  kind  may  he  take  to 
make  a  mixture  worth  26  cents  per  gallon  ? 

7.  A  trader  has  oils  at  $.95,  $1.20,  $1.42,  and  $1.60  per 
gallon,  of  which  he  wishes  to  ma^  a  mixture  worth  $1.25 
per  gallon.     How  many  gallons  of  each  kind  may  he  take  ? 

8.  A  trader  wishes  to  mix  50  lb.  of  sugar  at  7  cents 
per  lb.,  and  30  lb.  at  10  cents,  with  such  quantities  at  9  and 
6  cents  per  lb.  as  will  make  a  mixture  worth  8  cents  per  lb. 
How  many  pounds  of  each  may  he  take  ? 

9.  A  trader  wishes  to  mix  40  lb.  of  tea  at  40  cents  per  lb., 
SO  lb.  at  24  cents,  and  50  lb.  at  45  cents,  with  enough  at  30 
cents  to  make  a  mixture  worth  35  cents  per  lb.  How  many 
pounds  of  the  last  must  he  take  ? 

10.  I  have  salt  at  33,  37,  and  50  cents  per  bushel.  How 
many  bushels  of  each  kind  may  I  take  to  make  a  mixture  of 
100  bushels  worth  40  cents  per  bushel  ? 

11.  A  farmer  has  oats  worth  42  cents,  barley  worth  64 
cents,  rye  worth  87  cents,  and  wheat  worth  $1.38  per  bushel. 
How  many  bushels  of  each  kind  may  he  take  to  make  a  mix- 
ture of  200  bushels  worth  75  cents  per  bushel  ? 


SECTION    XV. 

INTEREST. 

lOG.    Introductory, 

When  a  person  hires  an  article  of  property  of  another,  it 
is  evident  that,  at  the  expiration  of  the  time  for  which  he 
hires  it,  he  ought  to  return  it,  and  pay  for  its  use.  Moreover, 
the  sum  paid  for  the  use  of  the  borrowed  article  should  be 
proportioned  both  to  its  value  and  the  length  of  time  it  is 
kept. 


INTEREST.  241 

For  instance,  if  I  hire  two  houses,  one  of  which  is  worth  twice  as 
mucn  as  the  other,  I  ought  to  pay  twice  as  much  per  year  for  the  first 
as  for  the  second.     If  the  values  of  the  houses  are  alike,  and  one  is 
kept  one  half  as  long  as  the  other,  only  one  half  as  much  ought  to  be  ^ 
paid  for  the  first  as  for  the  second.    ■ 

To  the  Teacher.  —  It  will  be  well  to  illustrate  the  above  imi)ortant  * 
principles  by  questions  similar  in  character  to  the  following  ;  — 

If  one  man  hires  a  horse  to  go  a  certain  distance,  and  another  lilies 
one  to  go  twice  as  far,  how  many  times  as  much  ought  the  second  to 
pay  for  its  use  as  the  first  pays  ?  "What  would  have  been  the  answer  to 
the  above  question,  provided  the  second  man  had  gone  3  times  as  far  as 
the  first  ?  4  times  as  far  ?  3^  times  as  far  ?  ^  as  far  ?  f  as  far "? 
§•  as  far  ?  &c.  If  the  horses  are  hired  by  the  hour,  and  the  first  man 
keeps  his  horse  three  times  as  many  hours  as  the  second  keeps  bis,  liow 
many  times  as  much  ought  he  to  pay  for  the  use  of  it  ?  What  would 
have  been  the  answer  had  he  kept  it  5  times  as  long  as  the  second  ?  8 
times  as  long  ?  6  times  as  long  ?  ^  as  long  ?  y  as  long  ?  ^  as 
long  ?   &c. 

Similar  questions  should  be  asked  with  reference  to  other  objects 
hired,  as  houses,  money,  &c.,  till  the  principle  is  fully  understood. 


107.    Definitions, 

{a.)  Money  is  very  frequently  hired,  and  the  sum  to  be 
paid  for  its  use  is  determined  in  accordance  with  the  above 
principles.     (See  177th  page,  Ex.  24,  Note.) 

(b.)  Money  thus  paid  for  the  use  of  money  is  called  In- 
terest. 

(c.)    The  money  used  is  called  the  Principal. 

(d.)  The  principal  and  interest  added  together  form  the 
Amount,  or  entire  sum  due  at  any  given  time. 

(e.)  The  interest  of  any  principal  is  usually  reckoned  at  a 
certain  per  cent,  i.  e.,  a  certain  number  of  one  hundredths  of 
that  principal,  for  each  year  it  is  on  interest.  This  per  cent 
is  called  the  Rate  per  cent,  or  simply  the  Rate, 

Note.  —  The  term  per  cent,  from  the  Latin  per  centum,  originally 
meant  by  the  hundred;  but  as  it  is  now  used  in  arithmetic,  it  means  one 
hundredtfis.     Thus  6  per  cent  means  -ffig-,  or  .06 :   4  per  cent  means 
21 


'242  INTEREST. 

r^(7,  or  .04  ;  ^  per  cent  means  x?T7,  or  ^  of  i^xj,  or  2^77;  &c.  This 
term  may  be  applied  to  any  thing  else  as  well  as  money ;  and  hence 
the  definition  (often  given  in  the  school  room)  "  so  many  cents  on  100 
cents  "  is  not  a  good  one,  any  more  than  would  be,  so  many  bushels  op 
100  bushels,  or  so  many  yards  on  100  yards.  It  is  the  more  objeclionablo 
because  scholars  are  sometimes  led  by  it,  and  by  being  often  called  upon 
to  use  the  term  per  cent  in  connection  with  money,  to  suppose  that  it 
has  some  necessary  connection  with  cents,  or  with  United  States  money 


1G8.    Legal  Rate. 

{a.)  In  most  countries,  laws  have  been  passed  regulating 
in  some  way  or  other  the  rates  of  interest. 

(h.)    Such  laws  are  called  Usury  Laws. 

(c.)    They  commonly  embrace  the  following  particulars  :  — 

First.  They  fix  the  rate  which  shall  be  paid  when  no 
special  rate  has  been  agreed  upon  by  the  parties.  This  is 
called  the  Legal  Rate. 

Second.  They  forbid  persons  to  receive  interest  at  more 
than  some  given  rate. 

Third.    They  impose  penalties  for  their  violation.* 

(rf.)  Any  excess  over  the  rates  allowed  by  these  laws  is 
called  Usury. 

(e.)  In  most  states  of  the  Union,  the  legal  rate  is  also  the 
highest  rate  allowed  by  law,  even  on  special  contracts.  The 
exceptions  to  this  are  named  in  the  following  statement  of 
the  legal  rates  of  interest  in  the  several  states. 

(/.)  In  New  York,  South  Carolina,  Georgia,  Michigan, 
and  Wisconsin,  the  legal  rate  of  interest  is  7  per  cent  per 
year. 

*  It  may  not  be  amiss  to  remark  that  the  laws  regulating  the  rate  of 
interest  are  very  often  disregarded,  while  the  penalties  for  their  violation 
are  seldom  imposed.  Very  few  men  continue  long  in  business  without 
paying  or  receiving  interest  at  more  than  the  legal  rate.  Money,  having, 
like  every  thing  else,  a  variable  value,  will  bring  what  it  is  worth  at  the 
time  it  is  sold  or  let,  and  it  seems  as  impossible  to  regulate  by  law  the 
price  which  shall  be  paid  for  its  use,  as  to  fix  by  law  that  which  shall  be 
paid  for  the  use  of  any  othci  article  of  property. 


INTEREST.  243 

(g.)    In  Alabama  and  Texas,  it  is  8  per  cent  per  year. 

(h.)    In  Louisiana,  it  is  5  per  cent. 

(^.)    In  California,  it  is  10  per  cent. 

(k.)  In  all  the  other,  states,  the  legal  rate  is  6  per  cent 
per  year. 

(I.)  By  special  agreement  of  the  parties,  interest  may  bs 
charged  at  the  rate  of  8  per  cent  per  annum  in  Florida,  Mis- 
sissippi, and  Louisiana ;  at  the  rate  of  10  per  cent  in  Arkan- 
sas, Elinois,  Michigan,  Iowa,  and  Ohio ;  at  the  rate  of  12 
per  cent  in  Wisconsin  and  Texas  ;  and  at  the  rate  of  1 8  per 
cent  in  California. 

(m.)  On  debts  in  favor  of  the  United  States,  interest  is 
computed  at  the  rate  of  6  per  cent. 

(n.)  In  each  state,  interest  is  reckoned  at  the  legal  rate  of 
that  state,  unless  otherwise  specified. 

(p.)  In  the  United  States,  it  is  customary,  when  reckoning 
interest,  to  regard  a  year  as  12  months,  and  a  month  as  30 
days.  But  in  England,  the  year  is  regarded  as  365  days,  and 
the  number  of  days  in  the  months  considered  are  reckoned  as 
in  the  calendar. 

(p.)    In  England,  the  legal  rate  is  5  per  cent. 

169.     Interest  for  2   Months,  200    Months,  ^c,  at    6  per 

cent. 

(a.)  If  the  interest  of  any  sum  for  one  year  is  6  per  cent 
of  that  sum,  for  ^  of  1  year,  or  2  months,  it  must  be  ^  of  6 
per  cent,  or  1  per  cent  of  it. 

(b.)  Therefore,  at  6  per  cent  per  year,  the  interest  for  2 
months  is  1  per  cent,  or  .01  of  the  principal,  and  may  be 
found  by  removing  the  decimal  point  of  the  written  principal 
two  places  towards  the  left. 

Thus,  the  interest  of  $75  for  2  months  is  $.75  ;  of  $364.30  is  $3,643, 
in.,  &C. 

What  is  the  interest  of  each  of  the  following  sums  for  2 
months  ? 


244 


INTEREST. 

1. 

$84 

3.     $827.41  * 

2. 

$687 

4.     $637.86 

$838.75 
$3.86 


7.  What  is  the  amount  of  each  of  the  above  sums  for  2 
months  ? 

(c.)  If  the  interest  of  any  sum  for  2  months  is  .01  of  that 
sum,  its  interest  for  .1  of  2  months,  which  is  6  days,  must  be 
,1  of  .01,  or  .001  of  it. 

(d.)  Therefore,  at  6  per  cent  per  year,  the  interest  for  6 
days  is  .001  of  the  principal,  and  may  be  found  by  removing 
the  decimal  point  three  places  to  the  left. 

Thus,  the  interest  of  $987  for  6  days  is  $.987  ;  of  $439  is  $.439 ; 
of  $8763.72  is  $8,764  ;t  or,  carrying  the  values  no  farther  than  cents, 
the  interest  of  $987  is  $.99 ;  of  $439  is  $.44 ;  of  $8763.72  is  $8.76  ;  &c. 

What  is  the  interest  of  each  of  the  following  sums  for  6 
days  ? 

8.  $586  10.     $67  12.     $1473.87 

9.  $930  11.  ^  $36.75  13.     $142 

14.  What  is  the  amount  of  each  of  the  above  sums  for  6 
days? 

(e.)  If  the  interest  of  any  sum  for  2  months  is  1  per  cent 
of  that  sum,  its  interest  for  100  times  2  months,  or  200 
months,  must  equal  100  times  1  per  cent,  or  100  per  cent  of 
it,  which  is  the  sum  itself. 

(/.)  Therefore,  at  6  per  cent  per  year,  the  interest  for  200 
months,  or  16  years  8  months,  must  equal  the  principal. 

Tims,  the  interest  of  $47  for  200  months  is  $47  ;  of  $37.84  is  $37.84  ; 
of  $23  is  $23  ;  &c. 

What  is  the  interest  of  each  of  the  following  sums  for  16 
yr.  8  mo.,  or  200  mo.  ? 


15. 

16. 


$38.73 
$57 


17. 

18. 


$67.95 
$2763 


19. 
20. 


$.28 
$1.07 


«  The  denominations  below  mills  need  not  be  given  in  the  answer.  In 
deed,  those  below  cents  need  nut  be  given,  if,  when  there  are  more  than  5 
mills:,  1  be  added  to  the  number  of  cents. 

t  Since  5.0^^072  =  more  than  i  of  a  mill. 


INTEREST.  245 

21.  What  is  the  amount  of^each  of  the  above  sums  for  200 
months  ? 

{g.)  If  the  interest  of  any  sum  for  200  months  is  equal  to 
that  sum,  its  interest  for  -^  of  200  months,  or  20  months, 
must  equal  y\j  of  it. 

(h.)  Therefore,  at  6  per  cent  per  year,  the  interest  for  20 
months,  or  1  year  and  8  months,  is  -^^  of  the  principal,  and 
may  be  found  by  removing  the  decimal  point  one  place  to  the 
left. 

Thus,  the  interest  of  $63  for  20  months,  or  1  yr.  8  mo.,  is  $6  30; 
of  $78.60  is  $7.86 ;  of  $8.79  is  $.879,  or,  carrying  the  result  only  to 
cents,  $.88 

What  is  the  interest  of  each  of  the  following  sums  for  20 
months,  or  1  yr.  8  mo.  ? 

22.  $73.86  24.     $673.70       i        26.     $5.87 

23.  $578  25.     $48.63         I        27.     $63 

28.  What  is  the  amount  of  each  of  the  above  sums  for  20 
months  ? 

170.    Recapitulation  and  Inferences, 

(a.)  The  importance  of  the  above  deductions  is  such  as  to 
demand  that  they  should  be  made  perfectly  familiar  by  all 
who  would  become  expert  in  casting  interest  at  6  per  cent. 
We  therefore  repeat  them. 

When  interest  is  6  per  cent  per  year,  — 

First.  The  interest  of  any  sum  for  200  months,  or  16  ^.  8 
mo.,  will  equal  that  sum. 

Second.  The  interest  of  any  sum  for  20  months,  or  \  yr,  8 
mo.,  will  equal  -^^  of  that  sum,  or  as  many  dimes  as  there  are 
dollars  in  the  principal. 

Third.  The  interest  of  any  sum  for  2  months  will  equal 
.01  of  that  sum,  or  as  many  cents  as  there  are  dollars  in  the 
principal. 

Fourth.  The  interest  of  any  sum  for  6  days  will  equal  .001 
of  that  sum,  or  as  many  mills  as  there  are  dollars  in  the  prin» 
cipal, 

21* 


246  INTEREST. 

{b.)  It  is  therefore  evident, that  for  ^  of  200  months,  the 
interest  will  equal  ^  of  the  principal ;  for  ^  of  200  months,  ^ 
of  the  principal ;  &c. 

(c.)  For  ^  of  20  months,  the  interest  will  equal  ^  of  -j^, 
or  2V  of  th®  principal ;  for  ^  of  20  mo.,  ^  of  i^jy,  or  ^\y  of  the 
principal ;  for  3  times  20  mo.,  3  times  .1,  or  .3  of  the  princi- 
pal ;  &c. 

(d.)  For  ^  of  2  months,  the  interest  wU  equal  ^  of  .01,  or 
jf^-^  of  the  principal ;  for  ^  of  2  months,  ^  of  .01,  or  ^^^  of 
the  principal ;  for  7  times  2  months,  7  times  .01,  or  .07  of 
the  principal ;  &c. 

(e.)  For  ^  of  6  days,  the  interest  will  be  ^  of  .001,  or 
jtjW  of  the  principal;  for  3  times  6  days,  3  times  .001,  or 
.003  of  the  principal ;  &c. 

ITl.     Table  showing  Interest  for  convenient  Fractional 
Parts  of  200  months,  20  months,  Sfc. 

At  6  per  cent  per  year,  the  interest  for  — 
200  mo.,  or  16  yr.  8  mo.,  =  prin. 

^  of  200  mo.,  or  100  mo.,  or  8  yr.  I  mo.,  =  j  of  prin. 

■g-  of  200  mo.,  or  66f  mo.,  or  5  yr.  6  mo.  20  da.,  =  ^  of  prin. 

i  of  200  mo.,  or  50  mo.,  or  4  yr.  2  mo.,  =  ^  of  prin. 

3-  of  200  mo.,  or  40  mo.,  or  3  yr.  4  mo.,  =  -^  of  prin. 

^  of  200  mo.,  or  33^  mo.,  or  2  yr.  9  mo.,  10  da.,  =  ^  of  prin. 

i  of  200  mo.,  or  25  mo.,  or  2  yr.  1  mo.,  =  ^  of  prin. 

lV  of  200  mo.,  or  20  mo.,  or  I'yr.  8  mo.,  =  tV  of  prin. 

iV  ol"  200  mo.,  or  I6f  mo.,  or  1  yr.  4  mo.  20  da.,  =  tV  of  prin. 
tV  of  200  mo.,  or  13^  mo.,  or  1  yr.  1  mo.  10  da.,  =  yV  of  prin. 
^  of  200  mo.,  or  12 j  mo.,  or  1  yr.  15  da.,  =  tV  of  prin. 

J-  of  20  mo.,  or  10  mo.,  =  i  o(  j\j  of  prin. 

^  of  20  mo.,  or  6f  mo ,  or  6  mo.  20  da.,  =  ^  of  iV  of  prin. 

^  of  20  mo.,  or  5  mo.,  =  i-  of  tV  of  prin. 

i  of  20  mo.,  or  4  mo.,  =«  -^  of  xV  of  prin. 

^  of  20  mo.,  or  3^  mo.,  or  3  mo.  10  da.,  =»  ^  of  -jV  of  prin. 

^  of  20  mo.,  or  2^  mo.,  or  2  mo.  1 5  da.,  =  i  of  -jV  of  pri(V 


INTEllEST. 


247 


t^  of  20  mo.,  or  2  mo.,  =  tV  of  tV  of  prin. 
■j^  of  20  mo.,  or  if  mo.,  or  1  mo.  20  da.,  =  yj  of  yV  of  prin. 
xV  of  20  mo.,  or  1^  mo.,  or  1  mo.  10  da.,  =         yV  of  yV  of  prin. 

^  of  2  mo.,  or  1  mo.,  =  ^  of  y^^  of  prin. 

^  of  2  mo.,  or  f  of  1  mo.,  or  20  da.,  =  ^  of  y  Ait  of  prin. 
:|-  of  2  mo.,  or  ^  of  1  rao.,  or  15  da.,  =          .       4"  of  y^  of  prin. 

■^  of  2  mo.,  or  f  of  1  mo.,  or  12  da.,  =  -g-  of  y^jy  of  prin. 

ij  of  2  mo.,  or  ^  of  1  mo.,  or  10  da.,  =  ^  of  y^cj  of  prin. 

yg-  of  2  mo.,  or  3"  of  I  mo.,  or  6  da.,  =  yV  of  y^^  of  prin 

y^^  of  2  mo.,  or  ^^  of  1  mo.,  or  4  da.,  =  yV  of  y^^  of  prin. 

^  of  6  da.,  or  3  da.,  =  ^  of  itjW  of  prin. 

■J^  of  6  da.,  or  2  da.,  =  ^  of  y^yW  of  prin. 

1^  of  6  da.,  or  I  da.,  =  ^  of  yuW  of  prin. 


179.     Questions  on  the  preceding  Table, 


1.  How  long  must  a  principal  be  on  interest  that  the  inter- 

est may  equal  ^  of  it  ? 

2. 

tV? 

15. 

i? 

27. 

rV? 

3. 

tV? 

16. 

iof^? 

28. 

tV? 

4. 

^of^? 

17. 

iofyV? 

29. 

iofyV? 

5. 

iofyV? 

18. 

tV?^ 

30. 

iofyV? 

6. 

fV? 

19. 

120  • 

31. 

^V? 

7. 

tV? 

20. 

^? 

32. 

yVofyV? 

8. 

xVofT^? 

21. 

?At7? 

33. 

Ti^? 

9. 

.01? 

22. 

\  of  .01  ? 

34. 

i  of  .01  ? 

10. 

1  of  .01  ? 

23. 

T^W? 

35. 

^n? 

11. 

fAh? 

24. 

S^UTS  • 

36. 

irAir? 

12. 

.001? 

25. 

■577077  • 

37. 

1  of  y^VxT  ? 

13. 

iroWr 

26. 

T^W? 

38. 

^  of  .001  ? 

14. 

V 

173.    Applications  of  the  foregoing  Table. 

1.  What  is  the  interest  of  $156.96  for  each  time  mentioned 
in  the  table  ? 


248  INTEREST. 

Answer.  —  The  interest  of  $156.96  for  200  mo.  =  $156.96  ;  for  100 
mo.,  or  8  yr.  4  mo.,  =  ;^  of  $156.96  ;  for  66f  mo.,  or  5  yr.  6  mo.  20  da., 
=  i- of  $156.96;  &c. 

The  interest  of  $156.96  for  20  mo.  =  $15,696  ;  for  10  mo.  =  ^  of 
$15,696,;  for  6f  mo.,  or  6  mo.  20  da.,  =  ^  of  $15,696  ;  &c. 

The  interest  of  $156.96  for  2  rao.  =  $1.57  ;  for  1  mo.  =  i"  of  $1.57  ; 
for  20  da.  =  ^  of  $1.57;  &c. 

Note. —  The  object  in  giving  the  answer  in  the  above  form  is  to 
direct  the  pupil's  attention  exclusively  to  the  method  of  computing  the 
interest;  but  as  soon  as  he  has  had  practice  enough  to  enable  him  to 
tell  at  once  what  part  of  the  principal,  or  of  some  convenient  part  of 
the  principal,  the  interest  for  any  of  the  above-mentioned  times  is,  he 
should  compute  the  interest  in  each  case.  Thus,  the  interest  of  $156.96 
for  100  mo.,  or  8  yr.  4  mo.,  =  3-  of  $156.96,  which  is  $78.48 ;  for  66f 
mo.,  or  5  yr.  6  mo.  20  da.,  =  -g-  of  $156.96,  which  is  $52.32  ;  &c. 

After  a  little  practice,  the  final  answer  may  be  read  at  once. 
Thus,— 

The  interest  of  $156.96  for  100  mo.,  or  8  yr.  4  mo.,  is  $78.48;  for 
66f  mo.,  or  5  yr.  6  mo.  20  da.,  is  $52.32 ;  &c. 

"What  is  the  interest  for  each  of  the  above-named  times  of, 

6.  $24.96? 

7.  $35.42  ? 


2.  $48  ? 

3.  $42.24  ? 


4.  $72  ? 

5.  $144? 

8.  What  is  the  interest  of  $36  for  3  yr.  4  mo.  ? 

9.  What  is  the  interest  of  $24.72  for  16f  mo.  ? 

10.  What  is  the  interest  of  $75  for  2  yr.  9  mo.  10  da.  ? 

11.  What  is  the  interest  of  $54  for  5  yr.  6  mo.  20  da.  ? 

12.  What  is  the  interest  of  $324  for  33^  rao.  ? 

13. .  What  is  the  interest  of  $231.12  for  1  yr.  15  da.  ? 

14.  What  is  the  interest  of  $42.24  for  2  yr.  1  mo.? 

15.  What  is  the  interest  of  $500  for  66§  mo.  ? 

16.  What  is  the  interest  of  $42.24  for  1  yr.  8  mo.  ? 

17.  What  is  the  interest  of  $150  for  6  mo.  20  da.  ? 

18.  What  is  the  interest  of  $4.80  for  5  mo.  ? 

19.  What  is  the  interest  of  $17.70  for  2  mo.  15  da.  ? 

20.  What  is  the  interest  of  $87.18  for  1  mo.  20  da.  ? 

21.  What  is  the  interest  of  $537  for  2  mo.  ? 

22.  What  is  the  interest  of  $288  for  10  da.  ? 


INTEREST.  249 

23.  What  is  the  interest  of  $96.34  for  20  da.  ? 

24.  What  is  the  interest  of  $675  for  2  da.  ? 

25.  What  is  the  interest  of  $438.74  for  3  yv.  4  mo.  ? 

26.  Wliat  is  the  interest  of  $73.87  for  1  jr.  15  da.  ? 

27.  What  is  the  interest  of  $144  for  1  yr.  1  mo.  10  da.  ? 

28.  What  is  the  interest  of  $dQ  for  4  yr.  2  mo.  ? 

29.  What  is  the  interest  of  $1728  for  1  yr.  4  mo.  20  da.  ? 

30.  What  is  the  interest  of  $327.29  for  16  yr.  8  mo.  ? 

31.  What  is  the  interest  of  $44.20  for  10  mo.  ? 

32.  What  is  the  interest  of  $57.60  for  3  mo.  10  da.  ? 

33.  What  is  the  interest  of  $43.65  for  4  mo.  ? 

34.  What  is  the  interest  of  $7.65  for  1  mo.  10  da.  ? 

35.  What  is  the  interest  of  $97.20  for  6  mo.  20  da.  ? 

36.  What  is  the  interest  of  $237.50  for  1  mo.  ? 

37.  What  is  the  interest  of  $541  for  5  da.  ? 

38.  What  is  the  interest  of  $9741  for  6  da.  ? 

39.  What  is  the  interest  of  $82.47  for  3  da.  ?    - 

40.  What  is  the  interest  of  $32.76  for  2  yr.  9  mo.  10  da.  ? 

41.  What  is  the  interest  of  $93.27  for  6  mo.  20  da.  ? 


174.    Interest  for  various  Times. 

(a.)  In  computing  interest  for  other  times  than  those  al- 
ready mentioned,  it  is  usually  most  convenient  to  divide  the 
time  into  parts,  as  illustrated  below. 

(b.)  The  student  should  bear  in  mind  that  in  the  forms  of  written 
work  here,  as  elsewhere,  the  letter  a,  b,  c,  &c.,  are  used  merely  to  indi- 
cate how  the  numbers  standing  opposite  them  have  been  obtained.  lu 
practical  work  they  should  be  omitted. 

1.   What  is  the  interest  of  $196.72  for  8  mo.  20  da.  ? 

First  Solution. 
a  =  $196.72    =  Principal. 


TjV  of  a  =  b  =        6.557  =  Int.  for  6  mo.  20  da 
.01  of  a  =  c  =        1 .967  =  Int.  for  2  mo. 

b  -f-  c  =  $    8.524  =  Int.  for  8  mo.  20  da. 


25C  INTERKST. 

Second  Solution. 
a  =  $196.72    =  Principal 

.04  of  a  =  b  =        7.869  =  Int.  for  8  mo. 
shu  of  a  =  c  =  .655  =  Int.  for  20  da. 


b  4-  c  =  $     8.524  =  Int.  for  8  mo.  20  da. 

Third  Solution. 
a  =  $196.72    =  Principal. 

2\j  of  a  =  b  =        9.836  =  Int.  for  10  mo. 
Y^jj  of  a  =  c  =        1.31 1  =  Int.  for  1  mo.  10  da. 

b  —  c  =  $     8.525  =  Int.  for  8  mo.  20  da. 

Fourth  Solution. 

Sini,e  at  6  per  cent  the  interest  for  1  month  equals  ^  of  1  per  cent 
of  the  principal,  the  interest  for  8  months  must  equal  4  per  cent  of  the 
princi])al ;  and  since  the  interest  for  1  day  equals  ^  of  .001  of  the  prin- 
cipal, the  interest  for  20  days  must  equal  ^^-  of  .001,  or  .003^  of  the 
principal.  Hence,  the  interest  of  $169.72  for  8  mo.  20  da.  =  .04  -f-  .003^ 
=  .043^  of  $169.72,  which,  found  by  the  usual  method,  gives  $8,525 
=  interest  for  8  rao.  20  da. 

2    What  is  the  amount  of  $649.37  for  17  mo.  15  da.  ? 

First  Solution. 
a  =  $649.37    =  Principal. 


-lVofa  =  b=      54.114  =  Int.  for  16mo.  20da. 
2^  *  of  b  =  c  =        2.705  =  Int.  for  25  da. 


a  +  b  +  c  =  $706,189  =  Amt.  17  mo.  15  da. 

Second  Solution. 

a  =  $649.37    =  Principal. 


iV  of  a  =  b  =      40.585  =  Int.  for  12  mo.  15  da. 
jVofa  =  c=      16.234  =  Int.  for  5  mo. 

a  -f-  b  +  c  =  $706,189  =  Amt.  for  17  mo.  15  da. 


•  Since  16  mo.  20  da.  =  600  da. 


INTEREST.  251 

Third  Solution.      " 
a  =  $649.37    =  Principal. 
lV  of  a  =  b  =      64.937  =  Int.  for  20  mo. 
^  of  b  i=  c  =        8.117  =  Int  for  2  mo.  15  da. 

a  +  b  —  c  =  $706,190  =  Amt.  for  17  mo.  15  da. 

Fowth  Solution. 
a  =  $649.37    =  Principal. 
2V  of  a  =  b  =      32.468  =  Int.  for  10  mo. 
j^  of  a,  or  2^  of  b,  =  c  =      16.234  =  Int.  for  5  mo. 
^V  of  a,  or  ^  of  c,  =  d  =        8.117  =  Int.  for  2  mo.  15  da 

a-f-b4-c  +  d=  $706,189  =  Amt.  for  17  mo.  15  da. 

Fijlh  Solution. 
Since  at  6  per  cent  the  interest  for  1  month  is  ^  of  1  per  cent  of  tho 
principal,  the  interest  for  17  months  must  equal  V-  of  1  per  cent  = 
.08^  =  .085  of  the  principal ;  and  since  for  1  day  the  interest  equals  ^ 
of  .001  of  the  principal,  for  16  days  it  must  equal  -^-  of  .001,  or  .002^ 
of  the  principal.  Hence,  the  interest  of  $649.37  for  17  mo.  15  da.  = 
.085  +  .002f  =  .087f  of  $649.37,  which,  found  by  the  usual  method, 
gives  $56,819,  the  interest,  which,  added  to  the  principal,  gives  the 
amount,  $706,189. 

Note.  —  Many  other  solutions  might  have  been  given  to  the  above 
questions,  but  as  they  would  all  involve  similar  principles,  it  is  unneces- 
sary to  add  them.  Every  question  in  interest  admits  a  great  variety  of 
solutions,  and  the  pupil  should  examine  it  carefully  to  determine  which 
he  will  adopt.  Practice  will  enable  him  to  select  a  good  method  at 
once.  One  process  may  be  applied  to  test  the  correctness  of  a  result 
obtained  by  some  other.  We  may  remark,  that,  as  a  general  thing,  it 
is  better  to  divide  than  to  multiply,  for  in  division  we  have  to  consider 
no  denomination  below  the  lowest  we  wish  to  have  in  the  answer. 

3.  What  is  the  interest  of  $857.63  for  3  mo.  16  da.  ? 

4.  What  is  the  interest  of  $875.37  for  1  mo.  26  da.  ? 

5.  What  is  the  interest  of  $93.75  for  9  mo.  29  da.  ? 

6.  What  is  the  interest  of  $178.43  for  16  mo.  14  da.  ? 

7.  What  is  the  interest  of  $343.65  for  13  mo.  16  da.  ? 

8.  What  is  the  interest  of  $237.64  for  19  mo.  24  da.  ? 

9.  What  is  the  interest  of  $478.96  for  17  mo.  26  da. 

10.  What  is  the  interest  of  $375.81  for  22  mo.  15  da.  ? 

11.  What  is  the  interest  of  $58.27  for  96  mo.  20  da.  ? 


252 


INTEREST. 


12.  What 

13.  What 

14.  What 

15.  What 

16.  What 

17.  What 

18.  What 

19.  What 

20.  What 

21.  What 

22.  What 

23.  What 

24.  What 

25.  What 

26.  What 


s  the  interest  of 
s  the  interest  of 
s  the  interest  of 
s  the  interest  of 
s  the  interest  of 
s  the  amount  of 
s  the  amount  of 
s  the  amount  of 
s  the  amount  of 
s  the  amount  of 
s  the  amount  of 
s  the  amount  of 
s  the  amount  of 
s  the  amount  of 
s  the  amount  of 


$5789  for  29  mo.  29  da.  ? 
$80.32  for  7  mo.  19  da.? 
$175  for  1  mo.  17  da.  ? 
$326  for  8  mo.  23  da.? 
$27.96  for  1  yr.  3  mo.  13  da.? 
$578.31  for  3  yr.  7  mo  28  da.  ? 
$724.16  for  7  yr.  2  mo.  11  da.? 
$4369.87  for  3  mo.  26  da.  ? 
$25.50  for  9  mo.  27  da.  ? 
$117.58  for  3  yr.  1  mo.  18  da.  ? 
$313.27  for  6  mo.  9  da.? 
$57.75  for  9  mo.  1  da.  ? 
$35.86  for  11  mo.  25  da.? 
$17.64  for  1  yr.  1  mo.  13  da.? 
51  for  1  yr.  5  mo.  17  da.  ? 


IT'S.     Computation  of  Time,  and  Application  to  Problems. 

{a.)  In  business  transactions,  it  is  usually  necessary  to 
compute  the  time  during  which  money  has  been  on  interest ; 
that  is,  the  time  between  the  dates  on  which  interest  began 
and  ended. 

(h.)  The  usual  method  of  doing  this  is  to  reckon  the  num- 
ber of  entire  years,  then  tlie  number  of  entire  calendar 
months  remaining,  and  then  the  remaining  days. 

(c.)  The  year  is  (as  before)  regarded  as  360  days,  or  12 
months  of  30  days  each,  and  each  entire  calendar  month  as  a 
month  of  30  days ;  but  the  days  which  are  left  after  reckoning 
the  years  and  montlis  are  determined  by  counting  them  ac- 
cording to  the  number  in  the  months  in  which  they  occur. 

1.  What  is  the  time  from  Jan.  17, 1845,  to  June  28, 1849? 
Solution.  —  From  Jan.  17,  1845,  to  Jan.  17,  1849,  is  4  years;  from 

Jan.  17  to  June  17  is  5  months;  from  June  17  to  June  28  is  11  days. 
Therefore  the  required  time  is  4  yr.  5  mo.  1 1  da. 

2.  What  is  the  time  from  April  27,  1846,  to  Feb.  13, 
1851? 


I 


INTEREST.  253 

Solution.  —  From  April  27,  1846,  to  April  27,  1850,  is  4  years;  from 
April  27,  1850,  to  Jan.  27,  1851,  is  9  months  ;  Januaiy  having  31  days, 
there  are  4  days  left  in  it,  which,  added  to  the  13  in  February,  give  17 
days.     Therefore  the  required  time  is  4  yr.  9  mo.  1 7  da. 

3.  What  is  the  time  from  Sept.  24,  1849,  to  March  20, 
1852? 

Solution.  —  From  Sept.  24,  1849,  to  Sept.  24,  1851,  is  2  years;  from 
bept.  24,  1851,  to  Feb.  24,  1852,  is  5  months;  1852  being  leap  year, 
February  has  29  days  ;  hence,  there  are  5  days  left  in  it,  which,  added 
to  the  20  in  March,  give  25  days.  Therefore  the  required  time  is  2  yr. 
5  mo.  25  da. 

Note. —  This  method  of  computing  the  time,  though  the  one  usual- 
ly adopted  by  business  men  when  interest  is  computed  for  months  and 
days,  is  unequal  in  its  operation  ;  for  the  calendar  months,  though  vary- 
•ing  in  length  from  28  to  31  days,  are  all  reckoned  as  months  of  30  days 
each.  Hence,  the  interest  of  a  sum  during  the  month  of  February  will 
be  as  much  as  during  either  March  or  April,  though  February  contains 
3  days  less  than  March,  and  2  less  than  April. 

By  this  method,  the  interest  on  four  notes  dated  respectively  on  the 
28th,  29th,  30th,  and  31st  of  any  one  month,  and  paid  on  any  one  day 
between  the  1st  and  28th  of  March,  of  any  year  except  leap  year, 
would  be  computed  for  the  same  time.  Suppose,  for  instance,  that  they 
are  dated  in  October,  1850,  and  paid  March  15,  1851.  Then  for  the 
first  note  dated  Oct.  28,  the  time  will  be  found  without  difficulty  to  be  4 
mo.  15  da.  In  calculating  the  time  on  the  others,  we  proceed  thus  : 
Since  there  are  not  as  many  as  29  days  in  February,  1851,  we  reckon 
from  the  29th,  30th,  or  31st  of  October  to  the  last  day  of  February  as  4 
months,  to  which  adding  the  15  days  in  March  gives  4  months  and  15 
days  as  the  time,  in  each  case.  The  restriction  with  reference  to  notes 
paid  in  leap  year  is  necessary  simply  because  February  has  then  29 
days.  The  proposition  will  always  be  true  of  notes  dated  on  the  29th, 
30th,  and  31st,  of  any  month,  and  paid  at  any  time  between  the  1st  and 
28th  of  March. 

Again.  Four  notes  dated  respectively  on  the  28th,  29th,  30th,  and 
31st  of  August  of  any  year  except  the  one  immediately  preceding  leap 
year,  and  payable  in  6  months,  would  all  become  due  on  the  same  day. 

The  only  strictly  accurate  method  of  reckoning  time  is  to  actually 
count  the  days  in  each  month  we  consider.  Thus,  to  reckon  the  time 
from  October  28,  1850,  to  March  15, 1851,  we  proceed  as  follows  :  From 
October  28th  to  3 1st,  is  3  days ;  to  which  adding  the  30  days  in  Novem- 
ber, the  31  in  December,  the  31  in  January,  the  28  in  February,  and  the 
15  in  March,  gives  138  days  as  the  true  time  between  the  two  dates. 
22 


254  INTEREST. 

In  England  the  time  is  al-?v^ays  computed  in  this  way,  as  it  is  in  thii 
country  when  notes  are  payable  at  the  end  of  a  certain  number  of  day- 

4.  What  is  the  time  from  June  23, 1850,  to  June  3, 1852  ? 
Answer.     1  yr.  11  mo.  11  da. 

5.  What  is  the  time  from  May  13,  1847  to  Oct.  8, 1851  ? 
-     6.   What  is  the  time  from  Jan.  31,  1851  .to  March  23, 

1852? 

7.  What  is  the  time  from  Nov.  17, 1849,  to  Dec  12, 1851  ? 

8.  What  is  the  time  from  Dec.  31, 1848,  to  July  6, 1850  ? 
What  is  the  interest  — 

^.  Of  $787.36  from  May  3,  1843,  to  Dec.  17,  1845  ? 

10.  Of  $54.76  from  Feb.  14,  1840,  to  June  2,  1844? 

11.  Of  $476.35  from  June  30,  1847,  to  Dec.  28,  1850  ? 

12.  Of  $638.29  from  May  31,  1851,  to  Oct.  7,  1852  ? 

13.  Of  $4937.56  from  Dec.  19,  1843,  to  Feb.  16,  1847  ? 

14.  Of  $481.74  from  Jan.  29,  1847,  to  March  25,  1851  ? 

15.  Of  $587.60  from  Jan.  31,-1850,  to  July  18,  1852  ? 
What  is  the  amount  — 

16.  Of  $947.84  from  May  15,  1850,  to  June  13,  1851  ? 

17.  Of  $748.67  from  Dec.  14,  1849,  to  May  4,  1851  ? 

18.  Of  $1546.61  from  April  9,  1847,  to  June  1,  1851  ? 

19.  Of  $917.68  from  June  5,  1842,  to  Jan.  1,  1850  ? 

20.  Of  $8396.58  from  April  30, 1847,  to  March  22, 1850  ? 

21.  Of  $1449.13  from  Dec.  31,  1850,  to  March  5,  1852  ? 

22.  Jan.  15,  1852,  George  W.  Pratt  borrowed  $237.50  of 
A.  N.  Johnson,  and  Feb.  13,  1852,  he  borrowed  $438.75 
more,  agreeing  to  pay  interest  at  6  per  cent  per  year.  He 
paid  the  debts  March  8,  1852.     What  was  their  amount  ? 

23.  I  have  three  notes  against  Arthur  Sumner,  viz.,  one 
for  $548.17,  dated  Jan.  1,  1851,  another  for  $679.18,  dated 
Jan.  27,  1852,  and  another  for  $376.89,  dated  May  31,  1852. 
What  amount  will  be  due  on  all  of  them  July  8,  1852  ? 

24.  Jan.  1,  1851,  I  borrowed  $3468,  with  which  I  pur- 
chased flour  at  $6  per  barrel.  I  sold  the  flour  March  17, 
1851,  for  $6.50  per  barrel,  cash.     Interest  being  reckoned  at 


INTEREST. 


255 


6  per  cent,  did  I  gain  or  lose  by  the  transaction,  and  how 
much  ?  . 

176.    Interest  hy  Days. 

(a.)  Many  business  men  always  reduce  the  time  to  days, 
and  compute  the  interest  by  the  method  illustrated  below.  As 
a  general  thing,  however,  this  method  is  not  so  convenient  as 
the  preceding. 

{b,)  Since,  at  6  per  cent  per  year,  the  interest  for  1  day  is 
i  of  txjVtt  of  the  principal,  it  follows  that  the  interest  for  any 
number  of  days  must  be  ^  as  many  thousandths  of  the  prin- 
cipal as  there  are  days. 

(c.)  We  may,  therefore,  find  the  interest  for  any  number 
of  days,  by  multiplying  the  principal  by  ^  of  the  number  of 
days,  and  removing  the  point  three  places  farther  to  the 
left. 

It  will  make  no  difference  with  the  result,  whether  we  multiply  by  ^ 
of  the  number  of  days,  or  by  the  number  of  days  and  divide  by  6  ;  but 
it  will  usually  be  better  to  divide  before  multiplying. 

1.   What  is  the  interest  of  $437.62  for  4  mo.  3  da.  ? 

Solution. —  Since  4  mo.  3  da.  =  123  da,,  the  required  interest  must 
be  ^  of  .123  of  the  principal,  which  is  .0202-  of  the  principal.    The 
work  carried  to  mills  would  be  written  thus  :  — 
$437.62 
.020j 

8.752 

.218 


$8,970  =  Arts. 

2.  What  is  the  interest  of  $54.57  for  4  mo.  20  da.  ? 

3.  What  is  the  interest  of  $397.42  for  8  mo.  15  da.  ? 

4.  What  is  the  interest  of  $231.48  for  7  mo.  12  da.  ? 

5.  What  is  the  interest  of  $438.64  for  5  mo.  24  da.  ? 

6.  What  is  the  interest  of  $281.87  for  5  mo.  9  da.  ? 

7.  What  is  the  interest  of  $581.21  for  6  mo.  24  da.  ? 

8.  What  is  the  interest  of  $83.25  for  2  mo.  21  da.  ? 

9.  What  is  the  interest  of  $98.37  for  6  mo  18  da.  ? 


256 


INTEREST. 


ITT.    Interest  hy  Dollars,  for  Months  and  convenient  Partt 
of  a  Month. 

(a.)  We  can  frequently  compute  interest  with  great  ease 
by  first  finding  the  interest  for  1  day,  1  month,  or  1  year,  and 
getting  the  required  interest  from  this. 

(6.)  When  the  interest  for  any  of  the  above  times  is  any  convenient 
sum,  as  1  dollar,  1  dime,  1  cent,  1  mill,  ^  dollar,  -^  dollar,  &c.,  the  pro- 
posed course  will  be  particularly  advantageous. 

(c.)  Since,  at  6  per  cent  per  year,  the  interest  of  any  sum 
for  1  month  is  -^j^  of  that  sum,  — 

1.  The  interest  of  $200  is  %1  per  month. 

2.  The  interest  of  $20  is  $.10,  or  1  dime  per  month. 

3.  2'he  interest  of  $2  is  $.01  per  month. 

1.   What  is  the  interest  of  $200  for  7  mo.  15  da.  ? 

Solution.  — The  interest  of  $200  is  $1  per  month;  therefore  for  7  mo. 
15  da.,  or  7^  mo.,  it  must  be  72"  dollars,  or  $7.50. 

What  is  the  interest  of  $200  for  — 


2. 

5  mo.? 

7. 

8  mo.  12  da.? 

3. 

8  mo.? 

8. 

9  mo.  15  da.? 

4. 

6  mo.  10  da.  ? 

9. 

1  yr.  7  mo.  ? 

5. 

1  yr.,  or  12  mo.  ? 

10. 

8  yr.  8  da.  ? 

6. 

2  yr.  1  mo.,  or  25  mo.  ? 

What  is  the  interest  for  each  of  the  above-mentioned 
times  — 

11.   Of  $2?         I      12.   Of  $20?       I      13.   Of  $2000? 

14.   What  is  the  interest  of  $100  for  7  mo.  ? 

Solution.  —  Since  the  interest  of  $200  is  1  dollar  per  month,  the  in- 
terest of  $100  must  be  a  half  dollar  per  month,  and  7  half-dollars,  or 
$3.50,  for  7  months. 

What  is  the  interest  of  $100  for  — 


15. 

9  mo.? 

20. 

8  mo.  15  da.? 

16. 

1  yr.? 

21. 

2  yr.  9  mo.  ? 

17. 

3  yr.  4  mo.  ? 

22. 

8  mo.  10  da.  ? 

18. 

15  da.  ? 

23. 

1  yr.  2  mo.  20  da.  ? 

19. 

20  da.  ? 

IN  fEREST. 


257 


What   is   the 
iimes — 

24.  Of  $10  ? 

25.  Of  $1  ? 


interest  for   each   of  the   above-mentioned 


27.  Of  $50  ? 

28.  Of  $5  ? 


26.   Of  $1000  ?         29.   Of  $500  ? 


30.  Of  $25  ? 

31.  Of  $250? 

32.  Of  $2500 .? 


ITS.     Interest   hy  Dollars,  when  the  Time  is  in  Days,  or 
Months  and  Days. 

(a.)  Since  the  interest  of  any  sum  for  6  days  is  .001  of 
that  sum,  the  interest  of  1  dollar  for  6  days  must  be  .001  of 
1  dollar,  which  is  1  mill.  If  the  interest  of  1  dollar  for  6 
days  is  1  mill,  its  interest  for  1  day  must  be  ^  of  1  miU. 
Therefore,  the  interest  of  $1  is  ^  of  1  mill  per  day. 

(b.)    From  this  we  have  the  following  statements  ;  — 

At  6  per  cent,  — 

1.  The  interest  of  $6  is  1  mill  per  day. 

2.  The  interest  of  $60  is  1  cent  per  day. 

3.  The  interest  of  $600  is  1  dime  per  day. 

4.  The  interest  of  $6000  is  1  dollar  per  day. 

Note.  —  When  the  interest  is  required  for  months  and  days,  we  may 
reduce  the  months  to  days,  and  then  apply  the  above  ;  or  we  may  find 
the  interest  for  the  months,  as  in  the  last  article,  and  then  find  it  for  the 
days,  as  above.  The  pupil  should  remember  that  the  interest  of  $6  is  3 
cents  per  month,  of  $60  is  3  dimes  per  month,  and  of  $600  is  3  dollars 
per  month. 


What  is  the  interest  of  $6  lor 


1. 
2. 
3. 
4. 
5. 
What 


15  da,  ? 

12  da.? 

20  da.  ? 

1  mo.  3  da.  ? 

3  mo.  10  da.  ? 

is   the   interest  for   each   of  the   above-mentioned 


7  mo.  15  da.  ? 

2  yr.  3  mo.  5  da.  ? 

1  yr.  4  mo.  11  da.  ? 

4  yr.  7  mo.  20  da.  ? 


times  — 

10.    Of  $60?       I       11.    Of  $600?     I       12. 
13.   What  is  the  interest  per  day  of  $750  ? 
22* 


Of  $6000  ? 


258 


INTEREST. 


Solution.  —  The  interest  of  $6000  being  1  dollar  per  day,  the  interest 
of  $750,  which  is  ^  of  $6000,  must  be  15  of  1  dollar  per  day. 

What  is  the  interest  per  day  — 


14. 
15. 
16. 
17. 
26. 
Bums  ? 


Of  $1000? 
Of  $2000  ? 
Of  $4000  ? 
Of  $1200? 
What  is  the  interest  per  month  of  each  of  the  above 


18. 

Of  $10? 

22. 

Of  $30? 

19. 

Of  $150? 

23. 

Of  $180  ? 

20. 

Of  $15  ? 

24. 

Of  $800  ? 

21. 

Of  $120? 

25. 

Of  $500  ? 

What  is  the  interest  of  $3000  for 


27. 

28. 
29. 
30. 
31. 


8  da.? 
15  da.? 

1  mo.  10  da.  ? 
20  da.  ? 

2  mo.  10  da.  ? 


32. 
33. 
34. 
35. 


27  da.  ? 
9  da.? 
25  da.  ? 
11  da.? 


36.   What  is  the  interest  of  $1200  for  each  of  the  above- 
mentioned  times  ? 


37.  Of  $100  ? 

38.  Of  $12? 

39.  Of  $.12  ? 


40.  Of  $120? 

41.  Of  $500  ? 

42.  Of  $1.20? 


43.  Of  $600  ? 

44.  Of  $250  ? 

45.  Of  $.60  ? 


1 70*     When  to  disregard  Cents, 

If  the  time  is  not  very  long,  the  interest  of  any  sum  less 
than  a  dollar  can  be  computed  with  sufficient  accuracy  by 
referring  the  principal  to  the  nearest  convenient  aliquot  part 
of  a  dollar. 

Thus,  the  interest  of  24  cents,  for  any  ordinary  time  of  calculating 
interest,  will  differ  but  a  trifle  from  that  of  25  cents,  or  ^  of  a  dollar, 
for  the  same  time ;  the  interest  of  35  cents  will  diflfer  but  a  trifle  from 
that  of  33 3^  cents,  or  ^  of  a  dollar,  &c. 

1.    What  is  the  interest  of  $599.77  for  9  mo.  17  da.  ? 

Solution.  — $599.77  =  $600  —  23  cents.  The  interest  of  $600  for  9 
mo.  17  da.  (being  $3  per  month,  and  1  dime  per  day)  is  $28.70.  As  23 
cents  is  very  near  25  cents,  its  interest  must  be  very  near  f  of  a  cent 
per  month,  which  will  be  about  1  cent  for  9  mo.  17  da.  This  taken 
from  $28.70  leaves  $28.69  as  the  interest  required. 


2.  What 

3.  What 

4.  What 

5.  What 

6.  What 

7.  What 

8.  What 

9.  What 
10.  What 


INTERKST.  259 

s  the  interest  of  $60.49  for  5  mo.  11  da.  ? 
s  the  interest  of  $59.67  for  8  mo.  13  da.  ? 
s  the  interest  of  $299.51  for  11  mo.  23  da.  ? 
s  the  interest  of  $150.32  for  7  mo.  19  da.  ? 
s  the  interest  of  $6.24  for  9  mo.  5  da.  ? 
s  the  interest  of  $200.42  for  8  mo.  15  da.? 
s  the  interest  of  $599.66  for  5  mo.  13  da.  ? 
s  the  interest  of  $119.94  for  13  mo.  7  da.  ? 
s  the  interest  of  $50.31  for  3  mo.  23  da.  ? 


180.     Interest  at   various  Rates,  obtained  from  that  at  6 
per  cent. 

When  the  interest  is  other  than  6  per  cent  per  year,  we 
may  first  find  the  interest  at  6  per  cent,  and  then  take  such 
part  of  this  as  the  given  rate  is  of  6  per  cent. 

Thus,  the  interest  of  any  sum  at  8  per  cent  is  f  •=  |^  =  1  ^  times  its 
interest  at  6  per  cent ;  a:  4^  per  cent  the  interest  is  —  =  f  of  the  in- 
terest at  6  per  cent,  &c. 

1.   What  is  the  interest  of  $367.32  for  1  yr.  9  mo.  20  da^ 

at  7^  per  cent  ? 

Solution. 

a  =  $367.32    =  principal. 


xV  of  a  =  b  =      36.732  =  int.  20  mo.  at  6  per  cent. 

3^2"  of  b  =  c  =        3.061  =  int.    1  mo.  20  da.  at  6  per  cent. 


b  +  c  =  d  =      39.793  =  int.  21  mo.  20  da.  at  6  per  cent. 
•^  of  d  =  e  =        9.948  =  int.  21  mo.  20  da.  at  l^-  per  cent. 


d  -f  e  =  $  49.741  =  int.    1  yr.  9  mo.  20  da.  at  72"  per  cent.  = 
Answer. 

Note.  —  The  work  may  frequently  be  facilitated  by  observing  that 
the  interest  of  any  sum  at  other  than  6  per  cent  is  equal  to  the  interest 
at  6  i»er  cent  of  the  same  part  of  that  sum  that  the  required  rate  is  of  6 
per  cent.  Thus,  the  interest  of  any  sum  for  a  given  time  at  3  per  cent 
is  equal  to  the  interest  of  f ,  or  ^,  of  that  sum  for  the  same  time  at  6 
per  cent.    The  interest  of  any  sum  at  4-^^  per  cent  is  equal  to  the  inte^ 

est  of  -^,  or  ^  of  that  sum,  at  6  per  cent,  &c. 


260  rXTERKST. 

Again.  The  interest  of  any  sum  at  other  than  6  per  cent  is  eqnal  to 
its  interest  at  6  per  cent  for  the  same  part  of  the  given  time  that  tho 
required  rate  is  of  6  per  cent.  Thus,  the  interest  of  any  sum  for  a 
given  time  at  2  per  cent  is  equal  to  its  interest  for  f ,  or  ^,  of  the  given 
time  at  6  per  cent. 

What  is  the  interest  of  — 

2.  $847.38  for  2  yr.  4  mo.  10  da.,  at  5  per  cent  ? 

3.  $483.94  for  3  yr.  5  mo.  26  da.,  at  7  per  cent  ? 

4.  $150  for  8  mo.  27  da.,  at  9  per  cent  ? 

5.  $46.38  for  11  mo.  19  da.,  at  3  per  cent? 

6.  $512.59  for  4  yr.  7  mo.  17  da.,  at  6^  per  cent  ? 

7.  $457.95  from  June  17,  1848,  to  May  19,  1850,  at  6J 
per  cent  ? 

8.  $978.31  from  Jan.  27,  1850,  to  Sept.  5,  1852,  at  5 
per  cent  ? 

9.  $87.63  from  April  21,  1848,  to  Jan.  7,  1852,  at  5^ 
per  cent  ? 

10.  $450  from  Aug.  12,  1852,  to  Oct.  7,  1852,  at  7^  per 
cent? 

11.  $75  from  Nov.  5,  1850,  to  Jan.  4,  1852,  at  3  per 
cent  ? 

12.  $240  from  Sept.  30,  1848,  to  May  23,  1851,  at  Ij 
per  cent  ? 

181.    Interest  at  various  Rates,  obtained  directly. 

Methods  similar  in  character  to  those  ilhistrated  in  the  fol- 
lowing examples  and  explanations  will  usually  be  more  brief 
than  the  preceding. 

1.   What  is  the  interest  of  $549.84  for  1  yr.  4  mo.  15  da., 

at  8  per  cent  ? 

Solution. 

a  =  $549.84    =  principal. 

jU8  of  a  =  b  =!  43.987  =  int.  for  1  yrr  at  8  per  cent 
^  of  b  =  c  =  14.662  =  int.  for  4  mo.  at  8  per  cent 
i  ot  c  =  d  =        1.833  =  int.  for  15  da.  at  8  per  cent 

b  -}-  c  +  d  =  $  60.482  =  int  foi  1  yr.  4  mo.  15  da.  at  8  per  cent  « 
Answer. 


INTEREST.  261 

Second  Solution.  —  Since  the  rate  is  8  per  cent  of  the  principal,  the 
interest  for  l:j-  years,  or  15  months,  must  be  1:^-  times  8  per  cent  =  10 
per  cent  =  xV  of  the  principal.    Hence, 
a  =  $549.84    =  principal. 


iV  of  a-  =  b  =      54.984  =  int.  for  15  mo.  at  8  per  cent. 

iV  of  b  =  c  =        5.498  =  int.  for  1  mo.  15  da.  at  8  per  cent. 


b  4-  c  =  $  60.482  =  int.  for  16  mo.  15  da.  at  8  per  cent  =  Ana, 

2.   What  is  the  interest  of  $537.48  for  3  yr.  8  mo.  15  da. 

at  7|  per  cent  ? 

Solution.  —  The  interest  for  2  yr.  at  7^  per  cent  mast  be  2  times  7|-, 
or  15  per  cent  of  the  principal.  Hence  we  have  the  following  written 
work :  — 

a  =  $537.48    =  principal. 


.15  of  a  =  b  =  80.622  =  interest  2  yr. . 

^ufb  =  c=  40.311  =  interest  1  yr. 

^  of  a  =  d  =  26.874  =  interest  8  mo. 

iV  of  d  =  e  =  1.679  =  interest  15  da. 


b-t-c4-d-f-e  =  $149,486  =  interest  3  yr.  8  mo.  15  da.  at  7j-  per 
cent. 

Second  Solution.  —  The  interest  for  4  yr.  at  1^  per  cent  must  be  4 
times  7^,  or  30  per  cent,  =  fV  of  the  principal.    Hence, 
a  :=  $537.48    =  principal. 

.3ofa  =  b=    161.244  =  interest  for  4  yr. 
fV  of  b  =  c  =      10.077  =  interest  for  3  mo. 
^ofc  =  d=        1.679  =  interest  for  15  da. 


b  —  c  —  d  =  $149,488  =  interest  for  3  yr.  8  mo.  15  da. 

3.   What  is  the  interest  of  $537.47  for  1  yr.  10  mo.  15  da. 
at  3^  per  cent  ? 

Solution. —  The  interest  of  any  sum  for  3  years  at  3^  per  cent  pel 
year  will  be  10  per  cent,  or  xV  of  that  sum.    Hence, 

a  =  $537.47    =  principal. 


2-V  of  a  =  b  =      26.873  =  int.  for  1  yr.  6  mo. 
^  of  b  =.c  =        6.718  =  int.  for  4  mo.  15  da. 


b  4-  c  =    $33,591  =  int.  for  1  yr.  10  mo.  15  da. 


262  INTEREST. 

Second  Solution. 
a  =  $537.47    =  principal. 


2V  of  a  =  b  =      26.873  =  int.  for  1  yr.  6  mo. 
i  of  b  =  c  =        4.479  =  int.  for  3  mo. 
^  of  c  =  d  =       2.239  =  int.  for  1  mo.  15  da. 


b  +  c  +  d  =    $33,591  =  int.  for  1  yr.  10  mo.  15  da. 

Note.  —  It  will  be  seen  that,  by  this  method,  we  first  get  the  ii  .est 
for  any  convenient  time,  and  then  take  such  part  or  parts  of  this  as  will 
give  the  interest  for  the  required  time. 

4.  What  is  the  interest  of  $279.64  for  6  yr.  5  mo.  22  da. 
at  ^  per  cent  ? 

Solution.  —  The  interest  for  4  years  at  -^  per  cent  must  be  4  times  ^, 
or  1  per  cent  of  the  principal.    Hence, 

a  =  $279.64    =  principal. 


.01  of  a  =  b  =  2.796  =  int.  for  4  yr. 

^  of  b  =  c  =  1.398  =  int.  for  2  yr. 
i|-*  of  b  =  d  =         .310  =  int.  for  5^  mo.,  or  5  mo.  10  da. 
■^■\  of  c  =  e=         .023  =  int.  for  12  da. 


$4,527  =  int.  for  6  yr.  5  mo.  22  da. 

"What  is  the  interest  of — 

5.  $483.79  for  3  yr.  3  mo.  18  da.  at  A}  per  cent  ? 

6.  $538.71  for  1  yr.  7  mo.  24  da.  at  9  per  cent? 

7.  $875.37  for  2  yr.  7  mo.  13  da.  at  8^  per  cent  ? 

8.  $63.29  for  5  yr.  11  mo.  10  da.  at  8^  per  cent  ? 

Suggestion.  —  The  interest  for  6  years,  at  8g-  per  cent,  is  50  per  cent, 
or  ^  of  the  principal,  and  since  20  days  =  -g-  of  2  months  =  iV  of  1 
year  =  t-J^  of  6  years,  the  interest  for  20  days  must  equal  -j-^^  of  the 
interest  for  6  years. 

9.  What  is  the  interest  of  $56.84  from  March  5,  1850,  to 
May  25,  1851,  at  5  per  cent  ? 

♦  Since  63-  mo.  =  ^  of  4  yr.,  or  48  mo. 

t  Since  12  da.  =  -^  of  2  mo.,  and  2  mo.  =  iV  of  2  yr.,  12  da.  must 
equal  ^  of  tV»  or  ^V  of  2  yr. 


INTEREST  f6$ 

10.  What  is  the  interest  of  $138.46  from  July  1,  1848,  to 
Aug.  26,  1852,  at  2^  per  cent  ? 

11.  What  is  the  interest  of  $273.81  from  Sept.  28,  1847, 
to  Oct.  31,  1852,  at  7  per  cent  ? 

12.  What  is  the  amount  of  $783.25  from  Aug.  28,  1846, 
to  Feb.  29,  1852,  at  4i  per  cent  ? 

13.  What  is  the  amount  of  $57.84  from  Jan.  19,  1849,  to 
Dec.  29,  1852,  at  6|  per  cent  ? 

14.  What  is  the  amount  of  $278.49,  from  Sept.  13,  1841, 
to  Oct.  10,  1846,  at  7^  per  cent  ? 

15.  Jan.  17,  1850,  I  borrowed  837  dollars,  agreeing  to  pay 
interest  at  the  rate  of  6  per  cent  per  year,  and  immediately 
put  it  on  interest  at  the  rate  of  7^  per  cent.  Aug.  27,  1852, 
I  collected  the  amount  due  to  me,  and  paid  that  which  I  owed. 
How  much  did  I  gain  by  the  transaction  ? 

Suggestion.  —  Since  I  paid  6  per  cent  and  received  7j-  per  cent  inter- 
est on  the  sum  I  borrowed,  my  gain  must  have  been  l^-  per  cent  per 
year. 

1 6.  A  merchant,  wishing  to  purchase  9  acres  of  land  at 
$378.43  per  acre,  borrowed  money  for  the  purpose  at  the  rate 
of  5  per  cent.  At  the  end  of  3  yr.  9  mo.  15  da.  he  sold  the 
land,  receiving  $400  per  acre  for  ^  of  it,  and  $475.28  per  acre 
for  the  remainder.     Did  he  gain  or  lose,  and  how  much  ? 

17.  Bought  397  yards  of  cloth  at  $3.75  per  yard,  payable 
in  6  months,  with  interest  at  7^  per  cent  per  year,  and  imme- 
diately sold  it  for  $4  per  yard,  payable  in  6  months,  without 
interest.  When  the  6  months  had  elapsed,  I  collected  the 
money  due  me,  and  paid  my  debt.  Did  I  gain  or  lose,  and 
how  much  ? 

18.  Bought  397  yards  of  cloth  at  $4  per  yard,  payable  in 
6  months,  and  immediately  sold  it  at  $3.75,  cash,  and  put  the 
money  on  interest  at  the  rate  of  7^  per  cent.  At  the  end  of 
6  months  1  called  in  the  money  I  had  lent,  and  paid  that 
which  I  owed.  Did  I  gain  or  lose  by  the  transaction,  ard 
how  much  ? 


264  PROMISSORY  NOTES. 

183.    Promissory  Notes. 

(a.)     A  PROMISSORY   NOTE,  a  NOTE   OP  HAND,  Or,  OS  it  is 

more  commonly  called,  a  note,  is  a  written  promise  to  pay  a 
specified  sum  of  money. 

(6.)  Annexed  is  a  form  of  a  note,  which,  with  the  subsequent  expla- 
nations, will  illustrate  the  principal  points  connected  with  this  subject. 

tJPo'i,  yo/lu^  lecBLuea,  Q)  kVom.t6e  to-  h/cui,  ^eaVoe 
c/ n^Lllv,  oV  oiaei,  orve  [viuva^o-  ooitaVi-j  wv  aem.cui/&,  lottn/  uv- 
teVe6t.  J^&Imi/     t^iou>iv. 

(c.)  The  above  note  is  a  promise  made  by  John  Brown  to  pay 
George  Smith,  or  order,  one  hundred  dollars,  and  is  equivalent  to  the 
following :  — 

Providence,  May  1,  1855. 
Because  of  an  equivalent  value  received  from  George  Smith,  I 
promise  to  pay  to  him,  or  to  whomsoever  he  may  order  me  to  pay  it, 
one  hundred  dollars^  whenever  the  payment  may  be  demanded  of  me 
by  presenting  this  note  ;  and  also  to  pay  the  interest  of  one  hundred 
dollars  from  this  date  till  the  time  of  payment 

John  Brown. 

(d.)   In  considering  this  note,  we  may  observe,  — 

First.  The  "  $100  "  placed  at  the  left  hand  upper  corner.  These  fig- 
ures do  not  form  an  essential  part  of  the  note,  but  arc  written  in  order  to 
enable  a  person  to  tell  at  a  glance  the  amount  for  which  it  was  given,  and 
also  to  guard  against  any  changes  which  might  be  made  in  the  body  of 
the  note. 

Second.   The  date,  which  shows  when  and  where  it  was  written. 

Third.  The  words  "  value  received,"  which  are  designed  as  an  ac- 
knowledgment that  the  signer  of  the  note  has  received  an  equivalent 
from  the  person  to  whom  it  is  to  be  paid. 

Fourth.  "  I  promise  to  pay  George  Smith,  or  order,"  which  means 
the  same  as,  "  I  promise  to  pay  to  George  Smith,  or  to  whomsoever  he 
may  order  me  to  pay  it." 

Fifth.  The  sum,  "  one  hundred  dollars."  This  should  always  bo 
rritten  out  in  words. 


PftOMTSSOKT    $iOTES.  265 

Sixth.  The  phrase  "  on  demand,"  which  means  whenever  he  shall 
demand  payment. 

Seventh.  The  phrase  "  with  interest,"  which  means  that  interest  is 
to  be  paid  from  the  time  the  note  is  dated. 

Eighth.  The  signature,  "  John  Brown,"  which  gives  validity  to  the 
note,  and  must  be  written  by  Mr.  Brown  himself,  or  by  some  one  special- 
ly authorized  by  him. 

{e.)  The  words  "  value  received "  are  regarded  as  essentikl  to  a 
note,  it  being  a  principle  in  law,  that  no  person  shall  be  compelled  to 
pay  money  for  which  he  has  not  received  an  equivalent  in  somts  lor  m  or 
other. 

(/)  The  words  "or  order"  may  be  omitte^d,  or  the  words  "or 
bearer  "  may  be  substituted  for  them.  If  they  are  omitted,  the  note  can 
only  be  collected  by  the  person  named  in  it.  Such  a  note  is  not  nego- 
tiable.    (See  183,  a.) 

ig.)  Some  specified  time,  as  "  in  sixty  days,"  "  in  three  months," 
&c.,  might  be  substituted  for  the  phrase  "  on  demand."  The  meaning 
then  would  be  in  so  many  days  or  months  from  the  date  of  the  note. 
Such  a  note  would  be  called  a  note  on  time. 

{h.)  Notes  on  time  are  not  regarded  as  due  till  three  days  after  the 
time  specified  in  the  note;  Thus,  a  note  payable  in  sixty  days  is  not 
due  till  the  end  of  sixty-three  days.  The  three  days  thus  added  are 
called  DAYS  of  geace. 

{{.)  If  grace  is  not  to  be  allowed,  the  form  of  the  note  should  be, 
"  in  so  many  days  or  months  without  grace." 

(j.)  If  the  last  day  of  grace  comes  on  Sunday,  or  upon  a  legal  holi- 
day, as  the  Fourth  of  July,  Thanksgiving,  &c.,  the  note  is  payable  on 
the  preceding  day,  and  if  that  be  Sunday,  or  a  legal  holiday,  it  is  pay- 
able on  the  first  day  of  grace, 

(k.)  If  the  phrase  "  with  interest"  should  be  omitted,  the  note  would 
not  be  on  interest.  If,  however,  a  note  on  demand  is  not  paid  when 
the  demand  is  made,  or  if  a  note  on  time  is  not  paid  when  due,  interest 
may  be  afterwards  charged,  though  no  mention  of  it  is  made  in  the 
note  ;  so  that  two  notes,  one  payable  in  three  months,  and  the  other 
payable  in  three  months,  with  interest  afterwards,  would  both  be  on  in- 
terest after  the  expiration  of  three  months. 

(/.)  The  form  of  notes  may  be  varied  somewhat  without  affecting 
ttieir  meaning  or  value.  Thus,  it  makes  no  difference  where  the  phrases 
*'  value  received "  and  "  on  demand"  are  placed,  provided  they  are 
Qbov»5  the  signature.  The  phrase  "  to  the  order  of  George  Smith" 
may  be  substituted  for  "  George  Smith  or  order."  These  and  other 
changes  will  be  illustrated  in  the  forms  of  notes  given  in  subsequent 
pnrvs  of  the  book. 

23 


Jf66  NEGOTIABLE   NOTES,    INDORSEMENTS,    BcC. 

ym.)  The  person  whose  name  appears  on  a  note  as  promisor  it 
called  the  maker  of  the  note. 

(w.)  The  person  to  whom  a  note  is  to  be  paid  is  called  the  i*atbb  or 
the  PROMISEE,  or  the  holder  of  the  note. 

(o.)  Every  note  should  be  written,  or  at  least  signed,  by  the  maker 
of  it,  or  by  some  one  specially  authorized  by  him.  It  is  then  taken  by 
the  payee,  and  is  his  property  until  it  is  paid,  or  until  he  transfers  it  to 
another.  When  paid,  it  should  be  given  up  to  the  signer,  who  should 
immediately  tear  off  or  erase  his  signature. 

(p.)  If  the  maker  of  a  note  refuses  or  neglects  to  pay  it,  when  the 
demand  is  legally  made,  at  ine  proper  place  and  time,  the  note  is  said  to 

be  DISHONORED. 

(q.)  A  demand  for  payment,  to  be  legal,  must  be  made  by  actually 
presenting  the  note ;  or,  if  it  is  payable  at  a  certain  place  on  a  given  day, 
by  having  the  note  deposited  at  that  place  ready  to  be  presented  to  the 
tnaker,  should  he  call  to  pay  it.  i 

Note. —  The  maker  of  a  note  is  not  obliged  to  regard  a  request  for 
its  payment  as  a  legal  demand,  unless  the  note  be  exhibited  and  ten- 
dered to  him  at  the  time  the  request  is  made ;  but  if  he  waives  his  rigit 
to  see  the  note,  the  demand  is  legal. 


183.    Negotiable  Notes^  Indorsements y  and  Protests. 

(a.)  A  NEGOTIABLE  NOTE  is  One  that  may  be  transferred 
or  sold  by  one  person  to  another. 

(6.)  Negotiable  notes  are  of  two  kinds,  viz.,  those  payable  to  a  per- 
eon,  or  order,  (as  "  George  Smith,  or  order, ^^)  and  those  payable  to  a  per- 
son, or  bearer,  (as  "  George  Smith,  or  bearer,")  or  simply  to  "  the  bearer" 

(c.)  Those  of  the  second  kind  are  negotiable  by  mere  delivery ;  but 
those  of  the  first  require  a  written  order  of  the  payee  to  authorize  any 
one  else  to  collect  the  money  due  on  them. 

{d.)  Such  an  order  is  commonly  written  on  the  back  of  the  note, 
and  is  called  an  indorsement. 

Thus,  if  George  Smith  wishes  to  transfer  the  above  note  to  Charlep 
Woods,  he  might  write  on  the  back  of  it,  "  Pay  to  Charles  Woods,  or 
order.  —  George  Smith." 

(e.)  This  would  be  an  indorsement  in  full,  and  would  give  Charlef 
Woods  the  same  title  to  the  note,  and  tlie  same  claim  on  John  Browr 
on  account  of  it  that  George  Smith  originally  had. 

{/.)   If  the   indorsement   hud    l)ecn,    "  Pay    to    Charles  Woods,  or 


/  NEGOTIABLE   N0TE6,   INDORSEMENTS,   &C.  ?67 

bearer,"  it  would  give  to  Charles  Woods,  or  whosoever  might  obtain 
legal  possession  of  tlie  note,  the  full  title  to  it. 

The  note  would  then  be  negotiable  by  mere  delivery. 

ig.)  If  it  had  been,  "  Pay  to  Charles  Woods  only,"  it  would  give  to 
Charles  Woods  the  full  title  to  the  note,  but  would  prevent  his  transfer- 
ring it  to  any  one  else. 

{h.}  If  George  Smith  had  simply  written  his  name  on  the  back  ot 
the  note,  it  would  have  been  an  indorsement  equivalent  to,  "  Pay  to  the 
bearer,"  and  would  make  the  note  negotiable  by  mere  delivery.  Such 
an  indorsement  is  called  an  indorsement  in  blank,  and  is  the  form 
most  frequently  used. 

(?.)  By  either  of  the  foregoing  forms  of  indorsement,  George  Smith 
would  not  only^iuthorize  some  one  else  to  collect  tlie  note,  but  he  would 
make  himself  responsible  for  several  important  points.  He  would 
guaranty,  — 

First    That  the  note  is  genuine,  and  just  what  it  purports  to  be. 

Second.   That  he  has  legal  possession  of  it,  and  a  ri<rht  to  transfer  it. 

Third.  That  it  shall  be  paid  if  payment  is  demanded  by  presenting 
the  note  to  the  signer  at  the  proper  time. 

Fourth.  That  if  not  so  paid,  he  will  pay  it  himself,  if  properly 
notified. 

If  he  does  not  wish  to  guaranty  the  last  two  points,  he  writes  the 
words  "  without  recourse  "  before  his  name.  The  indorsement  in  this 
form  is  a  guaranty  of  the  first  two  points,  but  not  of  the  last  two. 

(j.)  Any  person  may  write  his  name  as  a  special  indorser  on  a  note 
which  he  does  not  own.  Such  an  indorsement  would  not  aflfect  the 
negotiability  of  a  note,  but  it  would  make  the  indorser  responsible  for 
its  payment,  in  case  the  maker  should  not  pay  it.  Indeed,  it  mijrht  im- 
pose upon  him  all  the  obligations  with  regard  to  the  note,  which  rest 
upon  the  original  maker. 

(Ic.)  A  person  who  indorses  a  note  under  any  circumstances  incurs 
all  the  oblijrations  of  such  an  indorsement,  even  though  he  may  be 
ignorant  of  them  at  the  time.* 

{/.)  If  a  note  is  indorsed  by  several  persons,  each  of  them  makes 
himself  responsible  for  these  points  to  whoever  may  after\vards  get  legal 
possession  of  it. 

(w.)  A  person  to  whom  an  indorsed  note  is  transferred,  by  the  mere 
act  of  receiving  it,  agrees  with  all  the  indorsers.  except  the  special 
ones,  and  under  some  circumstances  with  them,  — 


*  Many  a  man  has  been  reduced  from  affluence  to  poverty,  by  merely 
writing  his  name  on  the  back  of  a  note  "just  to  accommodate  a  friend." 


26S  NEGOTIABLE    XOTES,    INDORSEMENTS,    &0. 

First.  That  payment  shall  be  demanded  of  the  signer  within  a  reason- 
able time,  if  the  note  is  payable  on  demand  ;  and  on  the  very  day  it  becotneM 
due,  if  it  be  on  time. 

Second.  That  he  will  not  consent  to  any  delay  in  the  time  of  pay- 
ment. 

Third.  That  if  the  note  is  dishonored  he  will  at  once  get  it  protest- 
ed, (see  (u.)  (v.)  p.  269,)  or,  on  that  day  or  the  next,  inform  the  indorser 
that  it  is  dishonoi'ed,  and  notify  him  that  he  will  be  held  responsible 
for  its  payment.  This  information  should  be  given  by  telling  him  per- 
sonally, or  by  leaving  a  letter  at  his  place  of  business  or  dwelling,  or  by 
mailing  one  to  his  address. 

(n.)  If  the  holder  of  a  note  fails  to  comply  with  any  one  of  these 
conditions,  he  releases  every  indorser  except  the  special  ones,  and  some- 
times even  them,  from  all  obligation  to  pay  it ;  but  he  does  not  iu  any 
way  affect  the  obligation  of  the  original  promisor. 

(o.)  The  question,  What  is  a  reasonable  time  in  the  case  of  a  note 
payable  on  demand  ?  must  depend  for  its  answer  somewhat  on  circum- 
stances. 

In  Massachusetts,  the  indorser  of  a  note  payable  on  demand  is  by 
statute  excused,  if  payment  be  not  demanded  within  sixty  days  from 
the  date  of  the  note. 

(p.)  If  no  particular  place  for  payment  is  specified  in  the  note,  it 
may  be  presented  at  the  signer's  counting  house  or  place  of  business,  in 
business  hours,  or  at  his  dwelling  house,  at  a  time  when  he  may  reason- 
ably be  supposed  to  be  at  home,  and  not  to  have  gone  to  bed. 

(7.)  If  it  is  payable  at  some  specified  place,  as  at  a  bank  or  at  some 
counting  room,  the  demand  must  be  made  at  that  place.  In  case  the 
promisor  does  not  appear  there  to  pay  the  note  on  the  day  it  falls  due, 
it  is  dishonored,  and  notice  should  be  sent  to  the  indorsers,  as  though 
payment  had  been  demanded  and  refused. 

(r.)  When  there  are  several  indorsers  to  the  note,  notice  that  it  is  dis- 
honored should  be  sent  to  each  whom  the  possessor  of  the  note  wishes 
to  hold  responsible  for  its  payment.  He  may  collect  the  money  of  either, 
or,  if  he  resorts  to  legal  measures,  may  commence  suits  against  any  of 
them,  or  against  all  at  once.  When,  however,  he  collects  the  money  of 
any  one,  his  demands  against  the  others  are  void. 

(s.)  Each  indorser  may  require  any  one  whose  name  precedes  his 
own  to  make  good  to  him  the  loss  he  may  sustain  on  account  of  the 
note,  provided  he  gives  notice  of  his  intention  to  do  so  the  day  that  he 
receives  his  own  notice,  or  the  day  after. 

For  instance,  suppose  that  the  foregoing  note  comes  to  me  indorsed 
by  George  Smith,  Charles  Woods,  Silas  Bacon,  and  Edward  Jones,  and 
thttt  it  is  dishonored.    I  shall  immediately  inform  Smith,  Woods,  Bacon, 


JOINT    AND    SEVERAL    NOTES.  269 

and  Jones  of  the  fact,  and  tell  each  that  I  shall  look  to  him  for  payment. 
Jones  will  on  the  same  day,  or  day  after  receiving  the  notice,  inform 
Smith,  Woods,  and  Bacon  of  it,  and  also  that  he  expects  them  to  make 
good  to  him  any  money  he  may  have  to  pay  on  account  of  it .  Bacon 
would  write  a  similar  letter  to  Smith  and  Woods,  and  Woods  would  write 
one  to  Smith ;  and  if  Smith  should  be  obliged  to  pay  it,  he  could  hold 
only  the  signer  of  the  note  responsible. 

(t.)  A  PROTEST  is  a  formal  declaration,  made  by  an  oflScer  called  a 
Notary  Public,  that  a  note  is  dishonored,  and  that  the  indorser  will  be 
held  responsible  for  it. 

(u.)  The  common  method  of  protesting  a  note  is  for  the  holder  to 
take  it  to  a  notary  public,  and  get  him  to  demand  payment  of  the 
promisor.  If  the  demand  be  refused,  the  notary  makes  out  a  protest, 
i.  e.,  he  writes  (usually  on  a  copy  of  the  note)  a  formal  declaration  that 
it  is  dishonored,  and  sends  a  copy  of  it  to  each  indorser.  The  protest 
should  be  made  out  on  the  very  day  the  note  is  due ;  otherwise  it  has  no 
value. 

(v.)  Although  a  notice  sent  by  the  holder  of  the  note  is  usually 
regarded  as  sufficient  to  make  the  indorsers  liable  for  its  payment,  it  is  a 
safer  course  to  have  a  formal  protest  made  out 


184: •    Joint  and  Several  Notes. 

(a.)  Notes  are  sometimes  signed  by  more  than  one  person.  If  the 
note  is  so  worded  as  to  show  that  the  signers  are  together  responsible 
for  its  payment,  it  is  a  "joint  note  ; "  but  if  so  worded  as  to  show  that 
the  signers  together  and  separately  are  responsible  for  its  payment,  it  is 
a  "joint  and  several  note." 

(6.)  There  seems  to  be  but  little  difference  between  a  joint  note  and 
a  joint  and  several  note,  as  far  as  regards  the  liabilities  of  the  promisors. 
Any  signer  to  either  may  be  compelled  to  pay  its  whole  amount.  The 
chief  difference  between  them  is,  that  an  action  commenced  to  enforce 
payment  of  a  joint  note  must  be  commenced  against  all  the  promisors 
jointly,  while  an  action  commenced  to  enforce  payment  of  a  joint  and 
several  note  may  be  commenced  against  them  jointly  or  individually. 

(c.)  If  judgment  is  given  in  favor  of  the  holder  of  a  joint  note,  he 
may  compel  either  promisor  to  pay  its  full  amount,  as  much  as  if  it  had 
been  a  joint  and  several  note.  One  who  is  thus  compelled  to  pay  the 
amount  of  the  note  may  recover  from  each  of  the  other  promisors  the 
share  which  each  ought  justly  to  pay.  In  a  joint  and  several  note,  or  a 
several  note,  he  may  or  may  not  recover  from  the  other  promisors,  ao 
cordmg  to  the  circumstances  under  which  the  note  was  given, 
23* 


270  RENEWAL    OF    NOTES. 


■«►     J^OAVitA-    HBlO/iR/,    0^     oloel,    pi|tit    doEEo/V^,    on,     demcux^,     uatK. 

WalteV    ^Je. 

(c?.)  The  above  is  a  joint  note,  because  it  is  the  promise  of  Sprague 
and  Hale  to  pay  jointly,  or  together,  a  specified  sum  of  money. 

(e.)  The  holder  of  the  note  may  demand  payment  of  either  Sprague 
or  Hale  ;  but  if  he  wishes  to  hold  any  indorser  or  iudorsers  responsible, 
he  must  demand  payment  of  both.  If  he  commences  an  action  on  ac 
count  of  it,  he  must  commence  it  against  both  jointly. 

{/■)  When  a  note  is  signed  by  one  person  as  principal,  and  another 
as  surety,  the  holder  of  it  may  demand  payment  of,  and  commence 
action  against,  the  surety  instead  of  the  principal,  if  for  any  reason  he 
chooses  so  to  do. 

185,    Renewal  of  Notes. 

(a.)  If  the  holder  of  a  note  allows  it  to  run  for  six  years  without  col- 
lecting any  thing  on  it,  or  otherwise  getting  it  renewed,  it  becomes  out- 
lawed, and  he  cannot  afterwards  compel  the  promisor  to  pay  it. 

(6.)  A  note  under  seal,  and  in  some  states  (as  Massachusetts  and 
Connecticut)  an  attested  note,  is  not  outlawed  under  twenty  years. 

(c.)  Any  act  by  which  the  promisor  acknowledges  the  validity  of  a 
note  renews  it.  Care  should  be  taken  to  have  the  renewal  made  in  the 
presence  of  witnesses,  or  to  have  evidence  of  it  in  the  handwriting  of 
the  promisor.    For  instance,  — 

{d.)  "When  a  part  of  the  money  due  on  a  note  is  paid,  a  receipt  for  it 
should  be  written  on  the  note.  Such  a  receipt  is  called  an  indorsement. 
Thus,  if  twelve  dollars  be  received  on  some  note,  the  following  would 
be  written:  "Jan.  1,  1855,  received  twelve  dollars."  This  indorsement 
should  be  written  by  the  one  who  pays  the  money,  as  it  is  equivalent  to 
a  renewal  of  the  note,  and  it  may  afterwards  be  important  for  the  holder 
to  prove  that  the  money  was  really  pai4  at  that  time  on  account  of 
tbe  note. 


FORMS    OF    NOTES. 


180.    Exercises. 


271 


^^00.  'Motion,,    ©t^Do^    i,    i^56. 

Q/n  Lrvetu.    dcutt^  a|tet    date,    0/     h/iom-Live    to-    ko/ii 

%am>ci>-    ^Veiu-,    ot    at^eV,    Pu>e  Tuuv^ie^    ^oEEo-i^.         ^UcOu/ft    Ve- 

celixed.  c/cunaeK     \o(vn,^n,. 

Who  is  the  maker  of  the  above  note  ?  To  whom  does  he  promise  to 
pay  it  1  Who  shall  take  possession  of  the  note  1  Can  he  transfer  it  ? 
If  so,  how  1  What  form  of  indorsement  would  be  used  if  he  wished  to 
transfer  it  to  Francis  Baker,  or  his  order  1  To  Francis  Baker  only  % 
To  Francis  Baker,  or  whoever  should  have  possession  of  it  ?  To  who- 
ever should  have  possession  of  it  without  naming  any  person  ?  How 
should  it  be  indorsed  in  each  of  the  above  cases,  if  the  indorser  does  not 
wish  to  make  himself  responsible  for  its  payment  ?  Is  it  on  interest  ?  On 
what  month,  and  what  day  of  the  month,  will  the  note  become  due  1 
If  this  note  should  be  indorsed  first  by  James  Drew,  then  by  Francis 
Baker,  and  then  by  William  Davis,  and  then  Charles  Morton  should 
come  into  legal  possession  of  it,  of  whom  and  when  ought  he  to  de- 
mand payment  1  If  payment  is  refused,  what  ought  he  to  do  1  What 
ought  each  indorser  to  do  on  receiving  notice  that  the  note  is  dishon- 
ored 1  Who  should  take  possession  of  it  when  it  is  finally  paid  1  What 
disposition  should  be  made  of  it  1  What  efifcct  would  it  have  on  the 
negotiaoility  of  the  above  note,  if  the  words  "  or  order  "  were  omitted  ? 
What  if  the  words  "  or  bearer  "  were  substituted  ? 

(1.) 


(Urv  aejYwuvOj  \Ay\m,  irvteVe6t  a|tei  tn/tce  rmyrvthA,  Cy 
Wix>tY\,L^  to  kcut  to  th/e  oloeV  ot  ^ok^enhM,  ^^'voaa,  Ptue  ruwvoVeo 
ooLLaV6 .       ^  (JcU/ite    Veceto  eo . 


^2  BANKS   AND    BANKING. 

(2.) 

^^  (3.) 

$800.  'MatR.,    ©(DM^u.^b    ^,    1^55. 

<To'l    0-cuKu*    ieceu^ca,   lo-e,  ^onri/     *^icHOft,  a5-    kiuv- 

cu|vaX,     tui/O-      \aunve,W     J^a/C«-60H',     cU>       ^u^etu,,     h/l-omi-^     to      twut 

©ouHi/'i-o-      ij%eea    etoftt    tuuvdieo-    0(H/tciA.6-,   orv     oeiTvcutaj    lulth/    wv- 


Let  the  scholar  write  each  of  the  above  notes  and  explain  their  mean- 
ing, and  the  meaning  of  all  their  points ;  let  him  also  change  their  form, 
and  indorse  ihem  in  various  ways. 

187,    Banks  and  Banking. 

(a.)  A  bank  is  an  institution  or  corporation  for  the  pur- 
pose of  trafficking  in  money. 

C^.)  Banks  usually  receive  money  on  deposit,  loan  money  on  interest, 
and  issue  bank  notes,  or  bank  bills,  i. «.,  notes  payable  in  specie  to  the 
bearer  on  demand  at  the  bank,  and  intended  to  circulate  as  money. 

(c.)  When  money  is  loaned  by  a  bank,  it  is  commonly 
made  payable  at  the  end  of  a  given  number  of  days,  and  the 
interest  for  that  time  and  the  three  days  of  grace  is  deducted 
at  the  time  it  is  borrowed. 

Thus,  on  a  note  of  $500,  payable  in  30  days.  I  shall  receive  at  a  bnnk 
|500,  minus  the  interest  of  $500  for  33  days.  i.  «<..  $500  —  $2.75  -  $497.25. 


HANKS    AND    BANKING.  273 

(d.)  By  this  arrangement  the  banks  receive  interest  on  a 
larger  sum  of  money  than  they  lend. 

Thus,  in  the  above  example,  the  bank  receives  interest  on  $500,  while 
the  sum  actually  lent  is  only  $497.25. 

(e.)  Bank  interest  is  called  discount,  because  it  is  thus 
deducted  from  the  face  of  the  note,  i.  e.,  from  the  sum  for 
which  the  note  is  given. 

(/.)    The  note  on  which  the  money  is  received  is  said  to 

be  DISCOUNTED. 

{g.)  To  present  a  practical  illustration  of  this  subject,  we  will  sup- 
pose the  following  case  :  — 

On  May  9,  1855,  George  Guild,  being  in  want  of  money,  wrote  a 
note  promising  to  pay  at  the  Merchants'  Bank,  to  the  order  of  Alfred 
Hall.  $600,  in  60  days,  and  got  Mr.  Hall  to  indorse  it.  He  then  applied 
to  the  officers  of  the  bank  to  discount  it,  and  they  decided  to  do  so.  He 
forthwith  presented  the  note  to  the  cashier,  who  deducted  the  interest  of 
$600  for  63  days,  and  paid  him  the  balance,  $59.3.70.  Guild  took  the 
money,  and  had  the  use  of  it  till  July  11,  when  the  note  became  due. 
He  then  paid  to  the  cashier  of  the  bank  the  $600  due  on  the  note,  and 
the  transaction  was  settled. 

{h.)  By  considering  the  above,  it  will  be  seen  that  Guild  paid  to  the 
bank  the  $593.70  which  he  had  borrowed,  together  with  the  interest  of 
$600 ;  so  that  he  paid  the  interest  of  $6.30  (i.  e.,  of  the  bank  discount) 
more  than  he  had  the  use  of. 

{{.)  If  he  had  wished  to  keep  the  money  as  much  longer,  he  would 
on  the  last  day  of  grace  have  written  a  new  note,  differing  from  tho 
former  only  in  the  date,  and  have  got  it  indorsed  as  before.  As  this 
new  note  would  be  worth  $593.70  at  the  bank,  he  could  by  giving  it,  and 
$6.30  besides,  to  the  cashier,  pay  the  amount  due  at  the  bank.  At  the 
end  of  63  days  he  would  again  owe  the  bank  $600. 

(j.)  Now,  it  is  obvious  that  during  all  this  time  he  has  been  paying 
the  interest  of  $600,  while  he  has  had  the  use  of  but  $593.70,  and  that 
therefore  he  has  paid  the  interest  of  $6.30  more  than  he  has  used.  Be- 
sides this  be  loses  the  use  for  63  days  of  the  $6.30  he  paid  on  renewing 
the  note.  Hence,  as  the  use  of  a  sum  is  worth  its  interest,  he  virtually 
pays  the  interest  of  $6.30  more  than  he  receives  for  126  days  -f-  63  days, 
or  1 89  days. 

(k  )  If  Mr.  Guild  should  fail  to  appear  at  the  bank  to  pay  the  note 
before  the  close  of  bank  hours  on  the  last  day  of  grace,  the  note  would 
be  protested,  and  notice  sent  by  a  notary  public  to  Mr.  Hall,  who  would 
then  be  held  responsible  for  its  payment. 


274  BANKtt    AND    BANKING. 

1.  A  note  of  $1200,  payable  in  60  days,  was  discounted  at 
a  bank  at  6  per  cent.     How  much  was  received  on  it  ? 

Solution.  —  The  interest  of  $1200  for  63  days,  beinr;:  2  dimes  per  day, 
IS  $12.60,  which,  deducted  from  $1200,  leaves  $1187.40  as  the  sum  re- 
ceived. 

2.  How  much  would  be  received  at  a  bank  on  a  note  of 
$200,  payable  in  90  days  ? 

3.  How  much  would  be  received  at  a  bank  on  a  note  of 
$360,  payable  in  30  days  ? 

4.  I  got  my  note  for  $1000,  payable  in  90  days,  discounted 
at  a  bank,  and  immediately  put  the  money  received  on  it  at 
interest.  When  the  note  became  due,  I  collected  the  amount 
of  what  I  had  put  on  interest,  and  paid  my  note  at  the  bank. 
How  much  did  I  lose  by  the  transaction  ?  How  does  the  sum 
lost  compare  with  the  interesl  of  the  bank  discount  for  the 
given  time  ? 

5.  My  note  for  $1000,  payable  in  6  months,  was  discount- 
ed at  a  bank,  and  I  immediately  put  the  money  received  on  it 
at  interest.  When  the  note  became  due,  I  collected  the  sum 
due  me,  and  paid  that  which  I  owed  at  the  bank.  How  much 
did  I  lose  by  the  transaction  ? 

6.  I  had  my  note  for  $500,  payable  in  2  months,  discount- 
ed at  a  bank,  and  immediately  put  the  money  on  interest. 
When  the  note  became  due,  I  renewed  it  for  the  same  time  as 
before;  and  when  the  new  note  became  due,  I  collected  the 
amount  due  me,  and  paid  my  note  at  the  bank.  How  much 
did  I  lose  ? 

Suggestions.  —  From  a  consideration  of  the  methods  of  reckoning  in- 
terest at  banks,  it  is  evident  that  from  the  time  the  first  note  was  dis- 
counted to  the  time  the  second  was  paid,  I  paid  interest  on  the  bank 
discount  more  than  I  received,  and  that  at  the  end  of  two  months  three 
days  I  paid  a  sum  equal  to  the  bank  discount.  Hence,  I  lost  the  inter- 
est of  the  bank  discount  for  4  mo.  6  da.,  plus  2  mo.  3  da.,  =  6  mo.  9  da. 

Or,  since  I  paid  nothing  at  the  bank,  except  the  bank  discount  at  the 
time  of  renewing  the  note,  and  the  second  note  when  it  became  due, 
the  actual  value,  at  the  time  of  settlement,  of  the  sums  paid,  will  be  the 
amount  of  the  bank  discount  for  2  mo.  3  da.,  plus  the  face  of  the  note 


ENGLISH   METHOD    OF    COMPUTING    INTEREST.  •      275 

The  sura  received  will  be  the  amount  for  4  mo.  6  da.  of  the  money  ob 
tained  at  the  bank  on  the  first  note.     The  difference  between  the  values  ^ 
paid  and  received  is  the  loss. 

7.  I  had  my  note  for  $600,  payable  in  4  months,  discounted 
at  a  bank,  and  immediately  lent  the  money  received  on  it  for 
just  one  year.  When  my  note  at  the  bank  became  due,  I  re- 
newed it  for  the  same  time  as  before,  and  when  this  new  note 
became  due,  I  renewed  it  for  such  time  that  it  became  due  at 
the  end  of  the  year,  when  I  collected  the  amount  of  the  sum 
1  had  lent,  and  paid  my  note  at  the  bank.  How  much  did  I 
lose  by  the  transactions  ? 

188.    English  Method  of  computing  Interest. 

(a.)  In  England,  time  is  reckoned  in  years  and  days,  but 
never  in  months.  The  year  is  regarded  as  365  days.  Inter- 
est is  usually  computed  by  first  finding  it  for  the  years,  and 
then  for  the  days. 

(ft.)  In  computing  it  for  the  days,  it  is  well  to  notice  that  73  days  = 
3^  of  1  year,  that  5  days  =  j-^  of  1  year,  and  that  1  day  =  -j^s  of  1 
year. 

(c.)  When  any  part  of  the  principal  is  expressed  in  shil- 
lings, pence,  and  farthings,  it  should  be  reduced  to  the  decimal 
of  a  pound.* 

*  Probably  the  simplest  method  of  doing  this  is  to  regard  each  shilling 
as  ^V,  or  .05  of  £1,  and  each  farthing  as  g^en,  or  .OOls-V  of  £1.  We  shall 
then  have  as  many  times  .05  of  £1  as  there  are  shillings,  plus  as  many 
times  .0015-4^  of  £1  as  there  are  farthings  in  the  pence  and  farthings.  But 
as  all  values  less  than  2"  of  .001  of  £1  are  so  small  that  they  may  be  dis- 
regarded, the  result  will  be  sufficiently  accurate  for  ordinary  purposes,  if 
we  regard  each  farthing  as  .001  of  £1,  observing  to  add  .001  if  there  are 
more  than  12  and  less  than  36  farthings,  and  .002  if  there  are  more  than 
36.  By  adding  this  result  to  the  value  of  the  shillings,  we  shall  have  the 
decimal  expression  required.  For  example :  To  find  what  part  of  £1  is 
equal  to  9  s.  8  d.  1  qr.,  we  have  9  s.  =  9  times  £.05  =  £.45 ;  8  d.  1  qr.  = 
33  qr.  =  £.033  +  £.001  =  £.034  Therefore,  9  s.  8  d.  1  qr.  =  £.45  +  £SiU 
=  £.484  The  reverse  operation  will  get  the  value  of  the  decimal  expres- 
sion, in  t«rms  of  shillings,  pence,  and  farthings. 


276       p:nglish  method  of  computing  interest. 

1.  What  is  the  interest  of  £327  17  s.  7  d.  from  May  7, 
185],  to  Sept.  4,  1852,  at  5  per  cent? 

Solution.  —  From  May  7,  1851,  to  May  7,  1852,  is  1  year.  There  are 
24  days  left  in  May,  to  which  adding  the  30  in  June,  the  31  in  July,  the 
31  in  August,  and  the  4  in  September,  gives  120  days.  The  time,  then, 
is  1  year  120  days.  The  principal  equals  £3271679  Hence  we  have 
the  following  written  work  :  — 

a  =  £327.879        =  principal. 


j05,  or  sV  of  a  =  b  =       16.39395    =  int.  for  1  yr. 
S^  of  b  =  c  =  .044914  =  int.  for  1  da. 

120  times  c  =  d  =         5.389680  =  int.  for  120  da. 


b  -f  d  =  e  =    £21.78363    =  int.  for  1  yr.  120  da. 
Or  we  may  have  the  following  :  — 

a  =  £327.879        =  principal. 


.05,  or  2V  of  a  =  b  =       16.39395    =  int.  for  1  yr. 
y^  of  b  =  c  =  .224574  =  int.  for  5  da. 

23  times  c  =  d  =         5.165202  =  int.  for  115  da. 


b-|-c  +  d  =  e=    £21.783726  =  int.  for  1  yr.  120  da. 

Calling  this  £21.784,  we  have  £21  15  s.  8  d.  1  qr.  as  the  answer. 
The  multiplications  required  in  solving  these  examples  render  it  neces- 
sary to  carry  out  the  work  to  places  below  thousandths,  though  we  do 
not  care  to  have  them  appear  in  the  answer. 

2.  What  is  the  interest  of  £47  9  s.  4  d.  1  qr.  from  May 

17,  1849,  to  Aug.  23,  1852,  at  5  per  cent? 

3.  What  is  the  interest  of  £148  19  s.  9  d.  3  qr.  from  Oct. 
23,  1850,  to  Nov.  11,  1852,  at  5  per  cent  ? 

4.  What  is  the  amount  of  £361   13  s.  2d.  1  qr.  from  July 

18,  1847,  to  April  12,  1850,  at  5  per  cent? 

5.  What  is  the  amount  of  £248  18  s.  10  d.  3  qr.  from 
Dec.  5,  1849,  to  March  3,  1852,  at  U  per  cent  ? 

6.  What  is  the  interest  of  £548  15  s.  7  d.  3  qr.  from  July 
29,  1847,  to  March  12,  1850,  at  2^  per  cent? 

7.  What  is  the  amount  of  ^258  19  s.  5  d.  2  qr.  from  Jan, 
I,  1849,  to  Sept.  29,  1852,  at  4  per  cent? 


PARTIAL    PAYMENTS.  277 

8.  What  is  the  amount  of  £329  7  s.  Id.  3  qr.  from  Nov. 
13,  1850,  to  Dec  1,  1852,  at  3  per  cent? 

9.  What  is  the  interest  of  £481  13  s.  5  d.  1  qr.  from 
April  19,  1842,  to  May  3,  1847,  at  5  per  cent  ? 

10.  What  is  the  amount  of  £222  2  s.  2  d.  2  qr.  from  Feb. 
29,  1848,  to  Jan.  1,  1852,  at  4  per  cent? 

180.    Partial  Payments. 

(a.)  The  principle  adopted  by  the  Supreme  Court  of  the 
United  States,  and  by  that  of  Massachusetts  and  most  of  the 
other  states,  as  the  one  to  be  applied  in  determining  the  sum 
due  on  a  promissory  note  or  bond  on  which  payments  have 
been  made  at  different  times,  is,  that  as  much  of  the  payment 
as  is  necessary  to  pay  the  interest  due  at  the  time  the  pay- 
ment is  made  should  be  appropriated  to  that  purpose,  and  the 
surplus  to  the  payment  of  the  principal.  The  balance  then 
due  will  form  a  new^  principal  on  interest  as  was  the  original 
principal.  If,  however,  any  payment  is  less  than  the  interest 
at  the  time  due,  the  principal  remains  unaltered,  and  on  inter- 
est, till  some  payment  is  made,  which,  with  the  preceding 
neglected  payments,  is  more  than  sufficient  to  pay  the  inter- 
est ;  when  we  proceed  as  if  a  single  payment,  equal  to  the 
sum  of  the  last  payment  and  the  preceding  neglected  ones  had 
been  made.* 

EXAMPLE. 

$750.00  Boston,  April  7,  1848. 

For  value  received,  I  promise  to  pay  James  Sullivan,  or 
order,  seven  hundred  and  fifty  dollars,  on  demand,  with  in- 
terest. Edward  Delano. 

*  The  method  adopted  hy  the  court  of  Connecticut  differs  from  the 
above  only  in  this  respect  —  that  if  a  payment  greater  than  the  interest  at 
the  time  due  be  made  before  the  principal  has  been  on  interest  one  year, 
the  person  making  it  is  allowed  interest  on  it  to  the  end  of  the  year ;  that 
b,  its  amount  from  the  time  it  was  made  to  the  end  of  the  year,  is  deduct- 
ed from  the  amount  of  the  principal  to  the  same  time.  If  settlement  oe 
made  before  the  principal  has  been  on  inteiest  one  year,  interest  is  allowea 
on  the  payments  from  the  time  thev  were  made  to  the  time  of  settlemenu 
24 


278  PARTIAL   PAYMENTS. 

On  this  note  are  the  following  indorsements  :  — 

Jan.  17,  1849.    Received  one  hundred  dollars. 

March  13,  1850.     Received  twenty-five  dollars. 

Feb.  19,  1851.     Received  thirty  dollars. 

Aug.  3,  1851.     Received  two  hundred  dollars. 

Jan.  1,  1852.     Received  one  hundred  and  fifty  dollars. 

What  was  due  on  the  note  at  the  time  of  settlement,  Aug. 
14,  1852  ? 

(6.)    The  following  exhibits  a  good  form  of  writing  the  work  in  such 
examples,  and,  in  connection  with  the  explanations  following,  will  be 
sufficient  illustration  of  the  process  :  — 

$750.00    =  1st  principal,  April  7,  1848. 
35.00   =  int.  280  days  at  1^^  dimes  per  day. 

$785.00   =  amt.  due  Jan.  17,  1849. 
$100.00    =  1st  payment. 


a 


$685.00   =  balance  due  Jan.  17,  1849. 

68.50    =  int.  20  mo. 

34.25    =  int.  10  mo. 

1.712  =  int.  15  da. 

.228  =  int.  2  da. 


2  yr.  6  mo.  17  da. 


$789.69    =  amt.  due  Aug.  3,  1851. 
255.00    =  2d,  3d,  and  4th  payments. 

$534.69    =  bal.  due  Aug.  3,  1851. 

13.367  =  int.  5  mo.    T      ^^  ___  ^         ^^  ^  ^^  ^9  da. 
.089  =  mt.  1  da.      I 


$547,968  =  amt.  due  Jan.  1,  1852. 
1 50.00    =  5th  payment. 

$397,968  =  bal.  due  Jan.  1,  1852. 
13.265  =  int.  6  mo.  20  da.  "| 
1.326  =  int.  20  da.  >   7  rao.  13  da. 

.198  =  int.  3  da.  J 


$412,757  =  amt.  due  Aug.  14,  1852,  =  Ans. 

Explanation.  —  As  it  is  obvious  that  the  first  payment  was  greater 
than  the  interest  at  the  time  due,  we  get  the  amount  of  the  note  to  that 
time,  and  deduct  from  it  the  payment.    The  remainder  is  a  new  prina- 


PARTIAL    PATMEXTS.  279 

pal  The  second  payment,  $25,  is  evidently  less  than  the  interest  then 
due ;  for  the  time  is  over  one  year,  while  the  principal,  between  $600 
and  $700,  gives  more  than  $36  interest  per  year.  Similar  considera- 
tions will  at  once  show  that  the  interest  to  the  time  of  the  third  pay- 
ment must  be  greater  than  the  second  and  third  payments  together. 
But  the  fourth  payment,  together  with  the  second  and  third,  is  very 
evidently  more  than  sufficient  to  pay  the  interest  then  due ;  therefore  we 
get  the  amount  of  the  new  principal  to  the  time  of  the  fourth  payment, 
and  subtract  from  it  the  sum  of  the  second,  third,  and  fourth  payments, 
thus  getting  our  third  principal.  As  it  is  evident  that  the  interest  of 
this  principal  to  the  time  of  the  fifth  payment  is  less  than  that  payment, 
we  find  its  amount,  and  from  it  subtract  the  payment.  This  gives  us 
the  fourth  principal,  which  is  on  interest  till  the  time  of  settlement ;  and 
hence  its  amount  is  the  sum  due. 

(c.)  The  above  is  really  equal  to  the  following  simple  problems,  each 
of  which  is  very  easy  of  solution  :  — 

Is  the  interest  of  $750  from  April  7,  1848,  to  Jan.  17,  1849,  more  or 
less  than  $100,  the  first  payment  1  What,  then,  is  the  amount  of  $750  for 
that  time  ?  How  much  will  be  due  after  paying  the  $100  ?  Is  the  in- 
terest of  this  to  March  13,  1850,  more  or  less  than  the  $25  at  that  time 
paid  ?  Is  the  interest  to  Feb.  19,  1851,  greater  or  less  than  $25  -f-  $30, 
or  $55,  the  sum  of  the  second  and  third  payments  ?  Is  the  interest  from 
Jan.  17,  1849,  to  Aug.  3,  1851,  greater  or  less  than  $255,  the  sum  of  the 
second,  third,  and  fourth  payments  ?  What,  then,  is  the  amount  of 
$685  from  Jan.  17,  1849,  to  Aug.  3,  1851  ?  How  much  will  be  due  after 
deducting  the  $255  ?  Is  the  interest  of  this  from  Aug.  3,  1851,  to  Jan. 
1,  1852,  greater  or  less  than  $150?  What,  then,  is  the  amount  of 
$534.69  for  that  time?  How  much  will  be  due  after  deducting  $150, 
the  fifth  payment"?  What  is  the  amount  of  this  from  Jan.  1,  1852,  to 
Aug.  14,  1852  1     What,  then,  was  due  Aug.  14,  1852  ? 

Every  problem  in  partial  payments  can  be  resolved  into  simple  ones ; 
and  if  the  pupil  will  use  a  little  care  in  determining  what  these  are,  and 
be  sure  that  he  performs  each  correctly,  he  may  obtain  a  true  result  the 
first  time  of  performing  the  work.  Nothing  short  of  this  should  satisfy 
him. 


2.       $850.00  g^g^^^^  ^p^j^  7^  J  3^7^ 

For  value  received,  I  promise  to  pay  Albert  Simmons,  or 
order,  eight  hundred  and  fifty  dollars,  on  demand,  with  in- 
terest. Isaac  Goodrich. 
On  thifc  note  were  the  following  indorsements :  — 
J  ne  19,  1848.     Received  one  hundred  and  twenty  five  dollars 


290  PARTIAL    PAYMENTS. 

Jan.  7,  1849.    Received  eighty-three  dollars. 
Sept.  27j  1849.     Received  one  hundred  dollars. 
May  1,  1850.    Received  twenty  dollars. 
Aug.  28,  1850.    Received  two  hundred  dollars. 
Jan.  1,  1851.    Received  one  hundred  dollars. 

How  much  was  due  on  the  note  Oct.  13,  1851  ? 


3.      $1000.00  Providence,  Not.  28,  1848. 

I  promise  to  pay  Bradford  Allen,  or  order,  one  thousand 
dollars,  on  demand,  with  interest.     Value  received. 

Henry  Williams. 

On  this  note  were  the  following  indorsements  :  — 

July  23,  1849.  Received  eighty  dollars. 

Feb.  28,  1850.  Received  fifteen  dollars. 

June  27, 1850.  Received  twenty  dollars. 

April  2,  1851.  Received  twenty-five  dollars. 

Dec.  20,  1851.  Received  five  hundred  dollars. 

May  17,  1852.  Received  three  hundred  dollars. 

How  much  was  due  Aug.  14,  1852  ? 


4.       $U5^  Worcester,  Dec.  20,  1846. 

For  value  received,  we  promise  to  pay  Alfred  Lincoln,  or 
order,  six  hundred  and  forty-five  dollars  and  seventy-fi»^« 
cents,  in  three  months,  with  interest  after. 

Thompson  &  French. 
Indorsements :  — 

Nov.  8,  1848.    Received  forty  dollars. 

April  16,  1849.    Received  three  hundred  dollars. 

March  10,  1851.    Received  two  hundred  and  fifty  dollars. 

Sept.  8,  1851.    Received  sixty  dollars. 

How  much  was  due  Jan.  1,  1852  ? 


5.       $1275.00  Bridgewater,  Sept.  29,  1845. 

For  value  received,  we  promise  to  pay  Lincoln  nnd  Wood 
twelve  hundred  and  seventy-five  dollars,  on  demand,  with  in- 
terest. Paine,  Root,  &  Co. 


PARTIAL    PAYMENTS.  281 

(ndorsements :  — 

Aug.  5,  1846.  Received  three  hundred  dollars. 

Sept.  22,  1847.  Received  four  hundred  dollars. 

May  25,  1848.  Received  two  hundred  dollars. 

June  17,  1849.  Received  one  hundred  and  fifty  dollars. 

Nov.  13,  1850.  Received  one  hundred  and  fifty  dollars. 

What  was  the  balance  due  March  1,  1851  ? 


6.       $3000.00  LoweU,  April  3,  1849. 

For  value  received,  I  promise  to  pay  the  order  of  James 

Wyman   three   thousand  dollars,  on   demand,  with    interest 

after  four  months.  -p,        -,  t^  i  • 

Jirdward  Kobmson. 

Attest,  George  Stone. 

Indorsements :  — 

Nov.  1,  1849.    Received  five  hundred  dollars. 
Dec.  27,  1850.     Received  ninety  dollars. 
March  25,  1851.     Received  fifty  dollars. 
July  18,  1851.     Received  six  hundred  dollars. 
Sept.  13,  1851.     Received  one  thousand  dollars. 
Jan.  1,  1852.     Received  one  thousand  dollars. 

The  note  was  settled  Nov.  8,  1852.      How  much  was  due  ? 


100.    Merchants^  Method  when  Debts   are  paid  within   a 

Year. 

(a.)  When  notes  or  debts  of  any  kind,  on  which  partial 
payments  have  been  made,  are  paid  in  full  within  one  year 
from  the  time  interest  commences,  merchants  often  determine 
the  sum  to  be  paid  on  settlement,  as  they  would  if  nothing  is 
due  on  a  note  till  it  is  paid  in  full ;  that  is,  they  find  the 
amount  of  the  note  to  the  time  of  settlement,  and  the  amount 
of  each  payment  from  the  time  it  was  made  till  the  time  of 
settlement,  and  then  consider  the  excess  of  the  amount  of  the 
note  over  the  sum  of  the  amounts  of  the  several  payments  to 
be  the  sum  due  on  settlement. 
24* 


282  PARTIAL    PAYMENTS. 


1.       $500.00  Worcester,  July  8,  1851. 

For  value  received,  I  promise  to  pay  John  F.  Barnard,  or 
order,  five  hundred  dollars,  on  demand,  with  interest. 

William  H.  West. 
Indorsements :  — 

Sept.  23, 1851.  Received  sixty  dollars. 

Nov.  20,  1851.  Received  one  hundred  dollars. 

Jan.  17,  1852.  Received  two  hundred  dollars. 

Feb.  8,   1852.  Received  fifty  dollars. 

How  much  was  due  May  11,  1852  ? 

Solution, 
$500.00  =  principal. 

25.25  =  int.  10  mo.  3  da. 


$525.25  =  arat.  of  note  to  May  11,  1852. 

$  60.00    =  1st  payment,  Sept.  23,  1851. 
2.28    =  int.  7  mo.  18  da. 
100.00    =  2d  payment,  Nov.  20,  1851. 

2.85    =  int.  5  mo.  21  da. 
200.00    =  3d  payment,  Jan.  17,  1852. 
3.80    =  int.  3  mo.  24  da, 
50.00    =  4th  payment,  Feb.  8,  1852 
.775  =  int.  3  mo.  3  da. 


$419,705  =  amt.  of  payments.  May  11,  1852. 


$106.55    =  bal.  due  May  11,  1852,  =  Arts. 


2.       $728.00  Springfield,  Sept.  7,  1849. 

For  value  received,  I  promise  to  pay  A.  Parish,  or  order, 
seven  hundred  and  twenty-eight  dollars,  on  demand,  with  in- 
terest. William  Mitchell. 

Indorsements :  — 

Oct.  3,  1849.    Received  eighty  dollars. 
Dec.  1,  1849.     Received  ninety  dollars. 
Feb.  4,  18.50.    Received  one  hundred  dollars 
March  2,  1850.    Received  forty  dollars. 
June  1,  1850.    Received  eighty  dollari. 

How  much  was  due  Aug.  2,  1850  ? 


PARTIAL    PAYMENTS.  283 


3.       $o83yViy  Lowell,  Jan.  18,  1850. 

For  valtie  received,  I  promise  to  pay  C.  C.  Chase,  or  order, 
five  hundred  and  eightj-three  dollars  and  seventy-five  cents, 
on  demand,  with  interest.  A.  H.  Fiske. 

Indorsements ;  — 

March  5,  1 850.     Received  fifty  dollars. 
April  1,  18.50.     Received  seventy-five  dollars. 
June  17,  1850.     Received  twenty-eight  dollars. 
July  3,  1850.     Received  one  hundred  dollars. 
Oct.  2,  1850.     Received  ninety  dollars. 
Dec.  27,  1850.     Received  seventy  dollars. 

How  much  was  due  Jan.  7,  1851  ? 


191.    To  find  the  Time. 

The  methods  illustrated  in  the  following  solutions  will  ena- 
ble us  to  find  the  time,  when  we  know  the  principal,  interest, 
and  rate. 

1.  How  long  must  $420  be  on  interest  at  6  per  cent  to 
gain  $32.27  ? 

Solution.  —  The  interest  of  $420  for  6  days  is  $.42  If  it  takes  6  days 
to  gain  $.42,  it  will  take  ^V  o^  6  days  to  gain  $.01,  and  3227  times  the 
last  result  to  gain  $32.27 

461 

^ff -  of  6  days  := days  =  461  days  =  15  mo. 

^^  ■  [H  days. 

2.  How  long  must  $357  be  on  interest  at  6  per  cent  to 
gain  $29,869  ? 

Solution.  —  The  interest  of  $357  for  6  days  is  $.357  If  it  takes  6 
days  to  gain  $.357,  it  will  take  -||f-  of  6  days  to  gain  $29,869 

251  2 

2||69  of  6  days  =  ^^^      ^  ^  days  =  502  days  =  16  mo. 
Z  t22aay. 


284  EQUATIOX    OF    PAYMENTS. 

3.  How  long  must  $136.80  be  on  interest  at  7  per  eent  to 
gain  $2,793  ? 

Solution.  —  The  interest  of  $136.80  for  1  year,  or  360  days,  at  7  per 
cent,  is  $9,576.  If  it  takes  360  days  to  gain  $9,576,  it  will  take  ||fB 
of  360  days  to  gain  $2,793 

21  5 

l^ff  01  360  da.  =  da,  r=  105  da.  ^  3  mo. 


Note.  —  By  this  method  of  solution,  we  first  select  some  convenient 
time  for  which  to  compute  the  interest.  Then  the  required  time  will  be 
the  same  part  of  the  time  selected,  that  the  given  interest  is  of  the  inter- 
est for  the  selected  time.  The  selected  time  should  be  one«for  which  the 
interest  can  be  easily  computed,  as,  when  the  rate  is  6  per  cent  per  year, 
6  da.,  60  da.  or  2  mo.,  600  da.  or  20  mo.,  6000  da.  or  200  mo. ;  and 
when  the  rate  is  other  than  6  per  cent,  1  year  =  360  da.,  or  such  part 
of  1  year  as  shall  give  for  the  interest  I  per  cent,  or  some  other  equally 
convenient  part  of  the  principal. 

When  there  is  a  fractional  part  of  a  day  in  the  result,  the  fraction 
may  be  omitted  if  it  be  less  than  ^ ;  but  if  it  be  more  than  ^,  1  may  be 
added  to  the  number  of  dftys. 

In  how  long  time  at  interest  will  — 

4.  $427.32  gain  $19.68  at  6  per  cent? 

5.  $186.75  gain  $12.45  at  6  per  cent  ? 

6.  $378.50  gain  $4,542  at  7  per  cent  ? 

7.  $56.34  gain  $18.78  at  8  per  cent? 

8.  $873.70  gain  $17,474  at  5  per  cent? 
a.  $594.00  gain  $00,654  at  4  per  cent  ? 


103.    Equation  of  Payments. 

{a.)  "When  one  man  owes  another  sums  of  money  payable 
at  different  times,  it  may  be  desirable  to  determine  when  the 
whole  can  be  paid  without  gain  or  loss  to  either  party.  The 
process  of  doing  this  is  called  equation  of  payments,  and 
the  time  sought  is  called  the  equated  time. 

{}).)  It  is  obvious  that  if  a  debt  be  not  paid  till  after  it  has 
become  due,  the  debtor  gains  the  use  of  it  from  the  time  it 


EQUATION    OF    PAYMKNTS.  285 

became  due  to  the  time  of  payment ;  while  if  it  be  paid  be- 
fore it  becomes  due,  the  debtor  loses  the  use  of  the  sum  paid 
from  the  time  of  payment  to  the  time  when  it  would  justly 
have  been  due.  The  use  of  any  sum  of  money  is  regarded 
as  worth  its  interest  for  the  time  it  is  used. 

(c.)  The  application  of  the  foregoing  principles  is  illus* 
trated  in  the  following  problems  and  solutions  :  — 

1.  Mr.  Lincoln  owes  Mr.  Wood  $400,  due  in  5  months, 
$600,  due  in  9  months,  and  $200,  due  in  12  months.  When 
can  the  whole  be  paid  without  gain  or  loss  to  either  party  ? 

Solution.  —  By  the  conditions  of  the  question,  Mr.  Lincoln  is  en- 
titled to  the  use  or 

Interest  of  $400  for  5  mo.  =  $10.00 
Interest  of  $600  for  9  mo.  =  $27.00 
Interest  of  $200  for  12  mo.  =  $12.00 


Or  to  use  $1200  till  its  int.  =  $49.00 
The  interest  of  $1200  being  $6  per  month,  he  is  entitled  to  keep  it  as 
many  months  as  there  are  times  $6  in  $49,  which  are  8^  times.     There- 
fore he  is  entitled  to  keep  it  8^  months,  or  8  months  and  5  days. 

First  Proof.  —  By  paying  the  whole  at  the  equated  time,  Mr.  Lincoln 
gains  the  use  of  the  first  debt  from  the  time  it  was  due  to  the  equated 
time,  and  loses  that  of  the  second  and  third  from  the  equated  time  to 
the  time  when  they  would  otherwise  have  been  due.     That  is,  he 

gains  interest  of  $400.00  for  3  mo.  5  da.    =  $6.33^ 

and  loses  interest  of  $600.00  for  25  da.  =  $2..50 

and  loses  interest  of  $200.00  for  3  mo.  25  da.  =  $3.83^ 


Sum  of  losses  =  $6.33^  =  the  gain, 
which  shows  the  work  to  be  correct. 

Second  Proof.  —  If  each  debt  should  be  paid  when  it  becomes  due, 
Mr.  "Wood  will,  when  the  last  debt  is  paid,  have  had  the  use  of  $400  for 
7  mo.  and  of  $600  for  3  mo.,  which  at  6  per  cent  is  equivalent  to  $14  -f- 
$9  =  $23  interest.  If,  however,  the  sum  of  the  debts  should  be  paid  at 
the  equated  time,  Mr.  Wood  would,  at  the  end  of  12  months,  when  the 
last  debt  would  otherwise  have  been  paid,  have  had  the  use  of  $1200  for 
3  mo.  25  da.,  which,  at  6  per  cent,  is  worth  $23  interest.  This  shows 
tliat  he  would  have  the  same  interest  in  one  case  as  in  the  other,  and 
thus  proves  the  first  result  correct. 

Second  Solution.  —  Another  solution  similar  in  character  to  the  last 
can  be  obtained  by  ascertaining  how  much  would  he  gained  or  lost  by 


286  EQUATION    OF    PAYMENTS. 

paying  the  entire  debt  at  any  assumed  time,  and  from  *hat  gtStting  th« 
equated  time. 

For  instance,  suppose  that  the  entire  debt  had  been  paid  at  the  end 
of  9  months.    Then  Mr.  Lincoln  would  have 

gained  interest  on  $400  for  4  months  =  $8.00 
and  lost  interest  on  $200  for  3  months  =  $3.00 

equivalent  to  a  gain  of  $5.00 
which  shows  that  9  months  is  as  many  days  longer  than  the  true  time 
as  it  will  take  for  $1200  to  gain  $5  at  interest.    We  find  (by  191)  that 
it  will  take  $1200,  at  6  per  cent  interest,  25  days  to  gain  $5.    Therefore 
the  equated  time  =  9  mo.  —  25  da.  =  8  mo.  5  da. 

Third  Solution.  —  When  the  numbers  are  convenient,  as  in  this  ex 
ample,  a  method  like  the  following  can  be  used  to  advantage  :  — 

The  sum  of  the  debts  is  $1200,  of  which  the  first  debt  is  ^,  the 
second  ^^  and  the  third  ^.  But  the  use  of  -3^  of  a  sum  5  mo.  is  worth 
as  much  as  the  use  of  the  whole  of  it  for  ^  of  5  mo.,  or  1§  mo. ;  the 
use  of  ^  of  a  sum  for  9  mo.  is  worth  as  much  as  the  use  of  the  whole 
of  it  for  2"  of  9  mo.,  or  4  J  mo. ;  and  the  use  of  ^  of  a  sum  for  12  mo.  is 
worth  as  much  as  the  use  of  the  whole  of  it  for  ^  of  12  mo.,  or  2  mo. 
Therefore  Mr.  Lincoln  is  entitled  to  the  use  of  the  sum  of  the  debts  for 
1§  mo.  -}-  42"  mo.  +  2  mo.  =  9>^  mo.  =  8  mo.  5  da. 

Fourth  Solution.  —  The  following  method  is  much  used,  but  we  think 
the  method  by  interest  will  ordinarily  be  found  preferable  :  — 

The  use  of  $400  for  5  mo.  is  worth  as  much  as  the  use  of  $1  for  400 
times  5  mo.,  or  2000  mo.  The  use  of  $600  for  9  mo.  is  worth  as  much 
as  the  use  of  $1  for  600  times  9  mo.,  or  5400  mo.  The  use  of  $200  for 
12  mo.  is  worth  as  much  as  the  use  of  $1  for  200  times  12  mo.,  or  2400 
mo.  Therefore  Mr.  Lincoln  is  entitled  to  the  use  of  the  entire  debt  for 
such  time  as  will  be  equivalent  to  the  use  of  $1  for  2000  mo.  -|-  5400 
mo.  -j-  2400  mo.,  or  9800  mo.  But  as  the  use  of  $1  for  9800  mo.  is 
equivalent  to  the  use  of  $1200  for  fs^crir  of  9800  mo.,  or  S^  mo.,  he  can 
keep  the  entire  debt  9>^  mo.,  or  8  mo.  5  da. 

Notes.  —  First.  As  the  equated  time  will  be  the  same,  whatever  be 
the  rate  of  interest,  the  rate  may  be  considered  to  be  that  which  is  most 
easily  calculated.  "  _ 

Second.  The  equated  time  will  frequently  contain  a  fraction  of  n 
day ;  but  if  the  fraction  be  less  than  J-,  it  may  be  disregarded,  or  if  it 
be  more  than  ^,  1  may  be  added  to  the  number  of  days. 

2.  A  owes  B  $250,  due  in  3  mo.,  $400,  due  in  6  mo.,  and 
$350,  due  in  8  mo.     What  is  the  equated  time  of  payment  ? 

3.  I  owe  $700,  payable  as  follows :    $150  in  3  mo.,  $184 


I 


TO    FIND    DATE    OF    EQUATED    TIME.  287 

in  7  mo.,  and  the  rest  in  1 1  mo.     When  can  1  pay  the  whole 
without  gain  or  loss  ? 

4.  I  owe  $960,  payable  as  follows  :  $180  in  4  mo.  20  da., 
$348  in  6  mo.  15  da.,  $234  in  8  mo.  5  da.,  and  the  rest  in  10 
mo.  13  da.     Required  the  equated  time  of  payment. 

5.  A  trader  bought  $1800  worth  of  goods,  agreeing  to  pay 
^  of  the  money  down,  ^  of  it  in  5  mo.,  ^  of  it  in  6  mo.,  ^  of 
it  in  9  mo.,  and  the  rest  in  12  mo.  At  what  time  may  the 
wliole  be  paid  ? 

6.  Bought  a  lot  of  goods,  for  which  T  agreed  to  pay 
$437.75  in  3  mo.,  $394.25  in  6  mo.,  and  $628.19  in  8  mo 
When  may  the  whole  be  paid  without  gain  or  loss  ? 

7.  A  owes  B  $800,  payable  in  10  mo. ;  but  to  accommo- 
date B,  he  pays  $250  down.  When  ought  the  remainder  to 
be  paid  ? 

Solution.  —  After  paying  $250,  he  will  owe  $800  —  $250  =  $550, 
which  he  ought  to  keep  till  its  interest  shall  equal  the  interest  of  $800 
for  10  months.  But  the  interest  of  $800  for  10  mo.,  equals  the  interest 
of  one  dollar  for  800  times  10  mo.,  or  8000  mo.  equals  the  interest  of 
$550  for  -^^^  of  8000  mo.,  which  is  14x\  mo.  Hence  it  ought  to  be 
paid  in  14inr  mo. 

8.  I  owe  $1000,  payable  in  9  mo. ;  but  to  accommodate 
my  creditor,  I  pay  $300  doAvn,  and  agree  to  pay  $300  more  in 
2  mo.     How  long  ought  I,  in  justice,  to  keep  the  remainder  ? 

9.  I  owe  $600,  payable  in  8  mo.  15  da.,  and  $400,  payable 
in  1 2  mo. ;  but  afterwards  agree  to  pay  $400  down,  and  $300 
in  2  mo.  20  da.,  on  condition  that  I  may  keep  the  remainder 
enough  longer  to  compensate  for  my  loss.  When  will  the 
remainder  become  due  ? 

10.  A  owes  B  $480,  due  in  1  yr.,  and  B  owes  A  $720, 
due  in  1  yr.  6  mo.  If  A  should  pay  his  debt  at  one  d,  when 
ought  B  to  pay  his  ? 

193.     To  find  Date  of  Equated  Time. 

(a.)  The  best  method  of  solving  such  examples  as  the  fol- 
lowing is  to  see  how  much  interest  will  be  gained  or  lost  by 
paying  the  sum  of  the  debts  at  any  assumed  time. 


288  TO    FIND    DATifi    OF    EQUATED    TIME. 

(6.)  It  will  be  well  as  a  general  thing,  to  select  for  the  assumed  time 
a  date  on  which  one  of  the  debts  becomes  due,  as  by  that  means  we 
shall  avoid  the  necessity  of  reckoning  interest  on  that  debt.  Reference 
should  also  be  had  to  the  probable  equated  time. 

(c.)  The  time  is  reckoned  by  counting  the  days  between 
the  dates  considered,  as  in  the  English  method  of  computing 
interest. 

1.  James  Brown  owts  "William  Greene  the  following  debts, 
viz. :  $534.83,  due  Jan.  7,  1855  ;  $285,  due  April  4,  1855  ; 
$327.38,  due  July  3,  1855  ;  and  $438,75,  due  Aug.  17, 1855. 
"When  may  the  whole  be  paid  without  gain  or  loss  ? 

Solution.  —  Suppose  that  Apul  4,  1855,  be  selected  as  the  assumed 

time.    Then  Mr.  Brown  would  gain  interest  on 

$534.83  from  Jan.  7  to  April  4,  88  da.      =  $7.84 

$285.00  due  at  assumed  time,  $0.00 

and  lose  int.  on 

$327.38  from  April  4  to  July  3,  90  da.  =  $4.91 
$438.75  from  April  4  to  Aug.  17,  135  da.  =  $9.87 

Sum  of  .  _  ^J5g5  gg  g^^  Qf  j^gggg  ^  $14.78 

debts       ) 

Excess  of  losses  over  gains $6.94 

Showing  that  Mr.  Brown  is  entitled  to  keep  $1585.96,  the  entire  debt 

due,  as  many  days  after  April  4  as  it  will  take  it  to  gain  $6.94  inter' 

est.    This,  found  by  191,  is  26  days,  plus  a  fraction  less  than  ^. 

Therefore  the   equated  time  is   26  days   after  April  4,  which  is 

April  30. 

Note.  —  The  above  shows  that  on  April  4th  Mr.  Brown  could  justly 
have  settled  the  account  by  paying  $1585.96  —  $6.94  =  $1579.02. 

Again.  Suppose  that  July  3  be  selected  as  the  assumed  time.  Then 
Mr.  Brown  would  gain  interest  on 

$534.83  from  Jan.  7  to  July  3,  178  da.,  =  $15.86 
$285.00  from  April  4  to  July  3,  90  da.,  =  4.27 
$327.38  due  at  assumed  time,  00.00 


Giving  for  sura  of  gains, 

$20.13 

and  lose  int.  on 

$438.75  from  July  3  to  Aug.  17,  45  da.,  = 
Sum  of  ) 

3.29 

debts. 


-«=  $1585.96  Excess  of  gain  over  loss,  $16.84 


TO    FIN»    DATE    OF    EQUATED    TIMK.  289 

•♦ 

Showing  that  Mr.  Brown  ought  to  pay  $1585.96,  the  entire  debt  due,  aa 
many  days  before  July  3  as  it  will  take  it  to  gain  $16.84  interest.  This, 
found  as  before,  is  64  days  nearly.  Therefore  the  equated  time  is  64 
days  before  July  3,  which  is  April  30,  as  before. 

Note.  —  The  above  shows  that  if  the  account  should  not  be  settled 
till  July  3,  Mr.  Brown  ought  justly  to  pay  $1585.96  -f-  $16.84  =* 
$1602.80. 

Proof.  —  By  paying  the  debt  ou  April  30,  Mr.  Brown  will  gain  in 
terest  on 

$534.83  from  Jan.  7  to  April  30,  114  da.=  $10.16 
$285.00  from  April  4  to  April  30,  26  da.  =      1.23 

Making  sum  of  gains  = $11.39 

He  will  lose  interest  on 

$327.38  from  April  30  to  July  3,  64  da.  =  $3.49 
$438.75  from  April  30  to  Aug.  17, 109da.=  $7.97 

Making  sum  of  losses,  = .  $11.46 

Excess  of  loss  over  gain,  = $00.07 

which,  being  less  than  the  interest  of  $1585.96  for  a  half  day,  shows 
that  April  30  is  the  correct  equated  time. 

2.  I  owe  $387.53,  due  Nov.  7,  1851  ;  $467.81,  due  Dec. 
21,  1851  ;  $256.19,  due  Feb.  11,  1852  ;  $136.43,  due  March 

I,  1852;  and  $387.59,  due  May  3,  1852.  What  is  the 
equated  time  of  payment  ? 

3.  I  owe  $2867,  due  April  15,  1850  ;  $1642,  due  July  27, 
1850;  $4371,  due  Oct.  8,  1850;  and  $5940,  due  Jan.  1, 
1851.     What  is  the  equated  time  of  payment  ? 

4.  I  owe  $628.13,  due  Dec.  17,  1852  ;  $427.19,  due  Dec. 
23,  1852  ;  $371.16,  due  Dec.  30,  1852  ;  $587.83,  due  Jan.  3, 
1853  ;  $987.62,  due  Jan.  7,  1853  ;  and  $843.28,  due  Jan.  14, 
1853.     What  is  the  equated  time  of  payment  ? 

How  much  is  due  on  the  above  Jan,  1,  1853  ? 

5.  I  owe  $543.28,  due  April  24, 1855  ;  $723.13,  due  May 

II,  1855;  $484,  due  Sept.  3,  1855;  $426.18,  due  Oct.  10, 
1855  ;  $236,  due  Nov.  10,  1855.  What  is  due  on  the  above 
Sept.  1,  1855,  interest  being  reckoned  at  5  per  cent  ? 

6.  Wliat  is  the  equated  time  for  paying  the  following 
debts  :  $600,  due  March  7,  1850  ;  $400,  due  June  11,  1850  ; 

25 


290  EQUATION  OP   ACCOUNTS. 

$800,  due  Aug.  17,  1850 ;    $500,  due  Oct.  3,  1850 ;   and 
$1000,  due  Nov.  27,  1850  ? 

194.    Equation  of  Accounts. 

{a.)  The  method  of  finding  the  equated  time  when  each 
party  owes  the  other,  that  is,  when  there  are  entries  on  both 
the  debit  and  credit  side  of  an  account,  does  not  differ  in  prin- 
ciple from  that  in  which  there  are  entries  only  on  one  side. 
The  following  example  and  solution  will  illustrate  it :  — 

1.   The  account  books  of  A  and  B  show  that 


•  A  owes  B 
$426.70,  due  Jan.  6,  1855. 
$413.65,  due  Feb.  2,  1855. 
$169.28,  due  April  13,  1855. 
$328.57,  due  Auf?.  29,  1855. 


And  that  B  owes  A 
$148.37,  due  Dec.  22,  1854. 
$173.19,  due  Jan.  29,  1855. 
$587,23,  due  May  7,  1855. 
$658.45,  due  Sept.  30,  1855. 


When  ought  the  balance  to  be  paid  ? 

Solution. —  Suppose  that  April  13, 1855,  be  the  assumed  time  of  pay- 
ment. Then  A  will  gain  interest  on  each  of  his  debts  which  becomes 
due  to  B  before  that  time,  and  on  each  of  B's  debts  which  become  due 
to  him  after  that  time ;  for  he  will  have  the  use  of  each  for  a  longer 
time  than  he  is  justly  entitled  to.  He  will  lose  interest  on  each  of  his 
debts  which  becomes  due  to  B  after  that  time,  and  on  each  of  B's  debts 
which  becomes  due  to  him  before  that  time ;  for  he  will  not  have  the 
use  of  them  for  so  long  a  time  as  he  is  justly  entitled  to.  Hence  A  will 
gain  the  interest  of 

$426.70  from  Jan.  6  to  April  13,  97  da.,  =  $  6.90 

$413.65  from  Feb.  2  to  April  13,  70  da.,  =  $  4.83 

$169.28  from  Feb.  13, $0.00 

$587.23  from  April  13  to  May  7,  24  da.,  =  $  2.35 

$658.45  from  April  13  to  Sept.  30,  170  da.,      =  $18.65 


Sum  of  gains  =     .    .    .  $32.73 

A  will  lose  the  interest  of 
$328.57  from  April  13  to  Aug.  29,  138  da.,  =  $7.56 

$148.37  from  Dec.  22,  1854,  to  April  13,  1855,  112  da.,  =  $2.76 
$173.19  from  Jan.  29,  to  April  13,  74  da.,  =  $2.14 

Sum  of  I0S.SCS, $12.46 

Excess  of  A's  gain  over  his  loss,  or  of  B's  loss  over  his  gain,       $20.27 


EQUATION'    OF    ACCOUNTS.  291 

But  the  sum  of  A's  debts  is  S1338.20,  and  of  B's  is  $1567.24, 
$1567.24  —  $1338.20  =  $229.04,  the  balance  which  B  owes  A. 

The  question  now  resolves  itself  into  this  :  If  by  B's  paying  A 
$229.04  April  13,  1855,  A  gains  and  B  loses  $20.27  interest,  when  can 
he  pay  it  without  any  gain  or  loss  of  interest  1  The  answer  evidently 
is.  As  many  days  after  April  13,  1852,  as  it  will  take  $229.04,  or,  disre- 
garding the  cents,  $229,  to  gain  $20.27  interest.  This,  found  by  methods 
before  explained,  is  531  days  =  1  yr.*  166  da.,  and  shows  the  equated 
time  to  be  Sept.  26,  1853,  which  may  be  proved  as  were  the  former  ex- 
amples. 

Note.  —  Although  accounts  like  the  above  are  sometimes  settled  by 
notes  payable  at  the  equated  time,  they  are  more  frequently  settled  by 
notes  payable  at  some  more  convenient  time,  or  by  cash.  In  all  such 
cases,  allowance  is  made  for  the  interest  gained  or  lost.  Thus,  if  the 
above  account  should  be  settled  by  cash  April  13,  1852,  $20.27  would 
be  deducted  from  the  balance  due  from  B  to  A,  in  order  to  compensate 
B  for  the  interest  he  would  lose ;  that  is,  B  would  pay  A  $229.04  — 
$20.27  =  $208.77.  If  it  should  be  paid  May  1,  1852,  B  would  have  to 
pay  A  $.69  (the  interest  of  the  balance  due  A  from  April  13  to  May  1) 
more  than  if  he  had  paid  it  April  13  ;  or,  which  is  the  same  thing,  he 
would  have  to  pay  the  balance  $229.04,  minus  its  interest  $19.58,  from 
May  1, 1852,  to  the  equated  time.  If  the  balance  due  at  any  given  time 
had  been  originally  required,  it  should  have  been  found  directly  by  mak- 
ing the  given  time  the  "  assumed  time." 

2.  By  the  respective  accounts  of  Heilry  Lane  and  William 
Pond,  it  appears  that 


Pond  owes  Lane 

$876.37,  due  April  5,  1852. 

579.48,  due  May  3,  18.52. 

487.83,  due  June  11,  1852. 

145.38,  due  Aug.  8,  1852. 


$2089.06  =  arat.  due  Lane. 


And  that  Lane  owes  Pond 
$228.13,  due  April  28,  1852. 

347.16,  due  June  3,  1852. 

313.27.  due  July  28,  1852. 

839.42,  due  Sept.  1,  1852. 


$1727.98  =  amt.  due  Pond. 


$2089,06  —  $1727.98  =  $361.08  =  balance  due  Lane. 
When  can  this  balance  be  paid  without  gain  or  loss  to 
either  party  ? 

Solution.  —  Suppose  it  to  be  paid  June  11,  1852.     Then  will   Mr 
Pond  gain  the  interest  of  — 

*  Reckoning  the  year  as  365  days,  as  is  always  done  in  such  cases,  un« 
less  it  includes  February  of  leap  year,  when  it  is  reckoned  as  366  days. 


292 


KQUATION    OF    ACCOUNTS* 


$876.37  from  April  5  to  June  11,  67  da.,  =  $  9.79 
579.48  from  May  3  to  June  1 1,  39  4a.,   =      3.77 

487.83,  due  June  11, =      0.00 

313.37  from  June  11  to  July  28,  47  da.,  =      2.45 
839.42  from  June  11  to  Sept.  1,  82  da.,  =    11.47 


Sum  of  gains  = 

He  will  lose  the  interest  of 

$145,38  from  June  11  to  Aug.  8,  58  da.,  =  $1.41 
228,13  from  April  28  to  June  1 1,  44  da.^  =  1.67 
347,16  from  June  3  to  June  11,  8  da.,    =        .46 


$27.48 


Sum  of  losses,      .    .  $  3.54 

Excess  of  gain  over  loss, ....  ..."  23.94 

As  Mr.  Pond  gains  this  interest  on  money  which  he  owes,  he  ought 
to  pay  the  debt  ($361.08,  the  balance  of  the  account)  as  many  days  be- 
fore June  11,  1852,  as  it  will  take  for  it  to  gain  $23.94  interest.  This 
gives  for  the  equated  time  398  days  before  June  11,  1852,  which  is  May 
10,  1851.  The  sum  necessary  to  settle  the  account  after  the  equated 
time  will  be  the  amount  of  the  balance,  $361.08,  from  the  equated  time 
to  the  time  of  settlement. 

3.   When  was  the  balance  of  the  following  account  due  ? 
Dr.        George  Ide,  in  account  with  James  Snow.       Cr. 


1849. 

1849. 

'~~' 

Jan.    17. 

To  Mdse.      . 

$336 

18 

Feb.      1. 

By  Mdse.      . 

$421 

30 

Jan.    31. 

To  Mdse.      . 

443 

17 

Feb.    27. 

By  Mdse.      . 

620 

00 

March  7. 

To  Cash,       . 

218 

63 

Mar.    13. 

By  Mdse.      . 

283 

17 

April  17. 

To  Mdse.      . 

500  00 

April  29. 

By  Mdse.      . 

482 

29 

May   28. 

To  Mdse.      . 

84 

36 

June      1 . 

By  Mdse.      . 

825 

13 

4.   When  was  the  balance  of  the  following  account  due  ? 
Dr.      George  Black  in  account  with  John  Brown.      Cr. 


1850. 

"1 

18,50. 

— — 

May    13. 

To  Mdse.  4  mo. 

$431 

17 

Juno      1. 

By  Mdse,  3  mo. 

$223 

62 

July    25. 

To  Mdse.  3  mo. 

256  38 

July      7. 

By  Mdse.  6  mo. 

150 

00 

Aug.    8. 

To  Mdse.  6  mo. 

431i72| 

July    22.!  By  Mdse.  3  mo. 

250 

00 

Si'-pt.  23. 

To  Mdse.  3  mo. 

585  41 

Sept.     1.  By  Cash, 

300 

00 

Nov.     7. 

To  Mdse.  3  mo. 

738  16 

Nov.    23.  By  Mdse.  2  mo. 
Dec.      l.|  By  Mdse.  3  mo. 

138;i6 
12231 

y 


TO    FIND    THE   PRINCIPAL,    OR    INTEREST. 


2S3 


Note.  —  4  mo.,  3  mo.,  &c.,  means  that  goods  were  sold  at  so  many 
months'  credit. 


5.   What  was  due  on  tlie  following  account  Jan.  1,  1853  ? 
Dr.      George  Mann  in  account  with  Henry  Guild.     Cr. 


1852. 

1852. 

— • 

Mav     5. 

To  Bal.  3  mo. 

$513 

43 

April  20. 

By  Mdse.  6  mo. 

$328 

13 

June  27. 

To  Mdse.  4  mo. 

624 

27 

May    10. 

By  Mdse.  3  mo. 

143 

27 

July     3. 

To  Mdse.  6  mo. 

831 

13 

June    13. 

By  Mdse.  4  mo. 

837 

19 

Sept.    5. 

To  Mdse.  2  mo. 

47 

62 

July      7. 

By  Mdse.  6  mo. 

56 

18 

Sept.  17. 

To  Mdse.  3  mo. 

125:53 

Aug.   20. 

By  Mdse.  4  mo. 

123'42 

Oct.    19 

To  Cash,       . 

387  j  00 

Oct.       1. 

By  Mdse.  3  mo. 

78  36 

Dec.     1. 

To  Cash,       . 

629  28 

Nov.    23. 

By  Mdse.  2  mo. 

12714 

6.  What  was  due  on  the  following  account  Jan.  1,  1853, 
interest  being  7  per  cent,  and  4  months'  credit  being  allowed 
on  each  entry  ? 

Dr.  David  H.  Daniels  in  account  with  Georofe  W.  Dean.  Cr. 


1852. 

1852. 

July     8. 

To  Mdse.      . 

$  236 

17 

July      3. 

By  Mdse.      . 

$439  27 

Aug.     1. 

To  Sundries, 

819 

63 

July    25. 

By  Mdse.      . 

213;16 

Sept.    4. 

To  Mdse.      . 

142 

^•^ 

Sept.   13. 

By  Mdse.      . 

100  00 

Nov.  13. 

To  Mdse.      . 

947 

22! 

Oct.     24.  By  Mdse.      . 

262|l8 

Dec.     8. 

To  Sundries, 

1050  00 

Nov.    30.  By  Mdse.      . 
Dec.    21.  By  Mdse.      . 

327  48 
520  75 

190.    To  find  the  Principal,  or  Interest,  from  the  Amount, 
Rate,  and  Time. 

(a.)  When  the  amount,  time,  and  rate  are  given  to  find  the 
principal  or  interest,  we  find  what  part  any  principal,  or  (if 
the  interest  be  required)  its  interest  for  the  given  time,  at  the 
given  rate,  is  of  its  amount,  and  then  take  this  part  of  th© 
given  amount. 

(b.)  The  first  step  towards  this  is  to  find  the  fraction  ex- 
pressing what  part  any  interest  for  the  given  time,  at  tba 
2.5* 


294  TO    FIND    THE    PRINCIPAL,    OR    INTEREST. 

given  rate,  is  of  its  principal.  This  fraction  will  alwayn  be 
the  same  part  of  the  given  annual  rate  that  the  given  time  is 
of  1  year,  or  360  days ;  or,  if  the  rate  is  6  per  cent,  it  will 
equal  the  fraction  expressing  the  part  which  the  given  time  is 
of  200  months,  or  6000  days.  The  amount,  of  course,  will 
equal  the  principal,  plus  the  fractional  part  of  it  which  the  in- 
terest equals. 

(c.)  Thus,  interest  for  1  yr.  7  mo.,  or  19  months,  at  6  per  cent  per 
year.;  =  siny  of  the  principal,  and  the  amount  =  ^%%  -\-  sW  =  ^hff 
of  the  principal.  Hence,  3-^  of  the  principal  =  24"7  of  the  amount, 
nnd  the  entire  principal  =  fy^,  and  the  interest  jV^g^,  of  the  amount 
jior  19  mo.  at  6  per  cent.  (The  same  fractions  would  have  been  obtained 
by  considering  the  interest  to  be  tI"  of  y^tr  of  the  principal.) 

{d.)  Again.  Interest  for  19  months  at  4^  per  cent  per  year  =  ^^ 
of  j^i  =  \%  of  24t7  =  -iuis  of  the  principal,  and  the  amount  =  f  §8- 
4-  ^y^  =  f  ^^  of  the  principal.  Hence,  the  principal  =  |f  ^,  and  tlie 
interest  =  ^Vt>  of  the  amount  for  19  mo.  at  A^  per  cent. 

{e.)  Again.  Interest  for  2  yr.  3  mo.  2  da.,  or  812  days,  at  6  per  cent 
per  year,  =  ^V^nr  =  i^^^tt^  of  the  principal,  and  the  amount  =  -f^S^ 
+  tVA  =  \t^%  of  the  principal.  Hence,  the  principal  =  ir^T.  ^^^ 
the  interest  =  ttW?  of  the  amount  for  2  yr.  3  mo.  2  da.  at  6  per  cent. 
(The  same  fractions  would  have  been  obtained  by  considering  the  inter- 
est to  be  %\%  of  T^iT  of  the  principal.) 

(/)  Again.  Interest  for  5  mo.  14  da.,  or  164  days,  at  7  per  cent  per 
year,  =  ^|4  of  -xlsTS  =  f^  of  t^tt  =  ^^V  of  the  principal,  and  the 
amount  =  %%%%  +  /A^tt  =  %%tst5  of  the  principal.  Hence,  the  prin 
cipal  =  f2"FTi  ^^^  the  interest  =  igrWr^  of  the  amount  for  5  mo.  14 
da.  at  7  per  cent. 

1.  What  principal  on  interest  at  6  per  cent  per  year  will 
amount  to  $884,125  in  1  yr.  2  mo.  10  da.  ? 

Solution.  —  Since,  at  6  per  cent  per  year,  interest  for  1  day  =  -^i^js 
of  the  principal,  interest  for  1  yr.  2  mo.  10  da.,  or  430  days,  must  equal 
^i^W,  or  ^\,  of  the  principal,  and  the  amount  must  equal  f  ^^  -|- 
■^ih^  or  f  M,  of  the  principal.  Hence,  ^^tj  of  the  principal  must  oquni 
f ^Tj,  and  the  principal  itself  must  ecjual  f  ^^  of  the  amount    ^yj  0^' 


TO    FIND    THE    PRINCTPAL,    OR    INTEirEST.  295 

$884,125  =  $825  =  principal    required.     The   mere   numerical  work 
may  be  indicated  thus  :  — 

1  yr.  2  mo.  10  da.  =  430  da. 

430      ^„     43      _i_600   643 

§4^^  of  $884,125  =  $825  =  principal  required. 

Second  Solution.  —  The  amount  of  $1  for  1  yr.  2  mo.  10  da.  =  $1.07^, 
and  if  one  dollar  is  required  to  gain  $1.07^,  as  many  dollars  will  be 
required  to  gain  $884,125  as  there  are  times  $1.07if  in  $884,125  But 
$884,125  -^  $1.07^  =  $884,125  -^  f^#  =  ff^  of  $884,125  =  $825,  as 
before. 

Proof.  — The  amount  of  $825  for  1  yr.  2  mo.  10  da.  is  $884,125 

2.  What  principal  on  interest  at  7  per  cent  per  year  will 
amount  to  $703,551  in  4  mo.  27  da.  ? 

Solution.  —  At  7  per  cent  per  year,  interest  for  4  mo.  27  da.,  or  147 
days,  =  -jf  ^  of  x^^j  =  t^^^tj  of  the  principal,  and  the  amount  = 
\M^u  H-  if  ^#Ty  =  tM^^  of  the  principal.  Hence  the  principal 
must  equal  ^f 3^4^  of  the  amount,  which  in  this  case  is  rff^^  of 
$703,551  =  $684.  The  mere  numerical  work  may  be  indicated  thus  :  — 
4  mo.  27  da.  =  147  da.    i|J  of  thtt  =  lUh- 

\Mi^  of  $703,551  =  $684  =  principal  required. 

Second  Solution.  —  The  amount  of  $1  at  7  per  cent  for  4  mo.  27  da. 
=  $1.02x2^;  and  if  one  dollar  is  required  to  gain  $1.02y^§,  as  many 
dollars  would  be  required  to  gain  $703,551  as  there  are  times  $1.02x2^  in 
$703,551.  But  $703,551  -f-  $1.02-l-f§  =  $703,551  -=-  iMH  =  jMii 
of  $703,551  =  $684,  as  before. 

Proof.  —  The  amount  of  $684  for  4  mo.  27  da.  at  7  per  cent  equals 
S703.551. 

What  principal  will  amount  to  —  * 

3.  $569,296  in  8  mo.  16  da.  at  6  per  cent? 

4.  $573.16  in  2  yr.  9  mo.  10  da.  at  6  per  cent  ? 
6.     $922.13  in  1  yr.  4  mo.  at  8  per  cent  ? 

6.     $378.82  in  1  yr.  4  mo.  20  da.  at  6  per  cent  ? 


296  DISCOUNT    AND    PRESENT   "WORTH. 

7.  $57.72  in  8  mo.  20  da.  at  10  per  cent  ? 

8.  $899,944  in  5  mo.  14  da,  at  6  per  cent? 


196.    Discount  and  Present   Worth, 

(a.)  Discount,  as  it  is  technically  called,  furnishes  the 
most  common  application  of  the  processes  of  the  preceding 
article  to  the  problems  of  business  life. 

(b.)  It  is  obvious  that  the  true  present  value  of  a  debt  due 
at  a  future  time  is  that  sum  of  money  which,  put  on  interest 
at  the  present  time,  will  amount  to  the  given  debt  at  the  time 
it  becomes  due. 

Thus,  when  the  rate  of  interest  is  6  per  cent  per  year,  a  debt  of  $106 
due  in  one  year  is  worth  the  same  as  a  debt  of  $100  due  now  ;  for  if  the 
money  received  on  the  second  debt  be  put  on  interest,  it  will  amount  to 
$106  in  one  year;  that  is,  when  the  first  debt  becomes  due. 

(c.)  A  debt  due  at  a  future  time  may  be  regarded,  then,  as 
the  amount  of  a  principal  on  interest  from  the  present  time 
to  the  time  when  the  debt  will  become  due.  This  principal 
is  usually  called  the  present  worth  of  the  debt,  and  its 
interest  is  called  the  discount,  because,  if  discounted  or 
deducted  from  the  debt,  it  leaves  the  present  worth. 

{d.)  From  this,  it  follows  that  to  ask  what  is  the  present  worth-of 
$651,  due  in  6  mo.  20  da.,  money  being  worth  6  per  cent  per  year, 
is  equivalent  to  asking  what  principal  on  interest  at  6  per  cent  will 
amount  to  $651  in  6  mo.  20  da. ;  and  that  the  solution  of  the  first  ques- 
tion is  the  same  as  that  of  the  second.  To  test  the  correctness  of  any 
result,  see  if  the  amount  of  the  present  worth  equals  the  given  debt. 

1.  What  is  the  present  worth  of  $438.18  due  in  1  yr.  6 
mo.  at  6  per  cent  per  year  ? 

Partial  Solution.  —  The  present  worth  required  is  that  sum  of  money 
which,  put  on  interest  at  6  per  cent,  will  amount  to  $438.18  in  1  yr.  6 
mo.,  or  18  mo.  The  interest  for  18  mo.  =  ^^^q-  =  t§^  of  the  princi 
pal,  and  the  amount  =,  &c.,  as  in  the  last  article. 

What  is  the  present  worth  of — 

2.     $83.45  due  in  4  yr.  2  mo.  at  6  per  cent  ? 
8.     $89.88  due  in  1  yr.  at  7  per  cent  ? 


DISCOUNT    AND    PRESENT    WORTH.  297 

4.  $142.56  due  2  yr.  hence  at  4  per  cent  ? 

5.  $122.94  due  4  mo.  27  da.  hence  at  6  per  cent  ? 

6.  $475.64  due  1  yr.  8  mo.  hence  at  6  per  cent  ? 

7.  $578.50  due  3  yr.  1^  mo.  hence  at  8  per  cent? 

8.  $731.52  due  3  yr.  4  mo.  hence  at  6  per  cent? 

9.  $1323.70  due  7  mo.  15  da.  hence  at  5^  per  cent? 

iO.  What  is  the  discount  of  $195.87  due  1  yr.  5  mo.  19 
da.  hence,  at  6  per  cent  per  year  ? 

Direction.  —  Find  what  part  of  the  debt  the  discount  is,  and  get  that 
part  of  $195.87.  For  proof,  subtract  the  discount  thus  found  from 
$195.87,  and  see  if  the  interest  of  the  remainder  for  the  given  time 
equals  the  discount.  "We  may  also  get  the  discount  by  finding  the 
present  worth,  and  subtracting  it  from  the  debt. 

What  is  the  discount  of — 

11.  $3946.11  due  2  yr.  5  mo.  15  da.  hence  at  6  per  cent? 

12.  $6392.43  due  15  mo.  7  da.  hence  at  6  per  cent  ? 

13.  $1241.27  due  1  yr.  5  mo.  23  da.  hence  at  6  per  cent  ? 

14.  $6255  due  3  yr.  2  mo.  hence  at  5  per  cent  ? 

15.  $179.96  due  2  yr.  3  mo.  6  da.  hence  at  4  per  cent? 

16.  I  own  a  note  for  $976,  payable  on  demand  with  in- 
terest, and  another  for  $1034.56,  payable  in  just  1  year,  with 
interest  afterward.  Allowing  money  to  be  worth  6  per  cent 
per  year,  which  debt  is  justly  worth  the  most  at  the  present 
time,  and  how  much  the  most?  Which  will  be  worth  the 
most  at  the  end  of  the  year  ?  Which  will  be  worth  the  most 
at  the  end  of  6  months  ?  Which  will  be  worth  the  most  at 
the  end  of  2  years  ? 

17.  If  two  notes  are  given  on  the  same  day,  one  for  a  cer- 
tain sum  due  at  a  future  time,  with  interest  afterwards,  and 
the  other  for  the  true  present  worth  of  the  first  note,  payable 
on  demand  with  interest,  their  true  values  will  be  the  same  at 
the  time  they  are  given,  and  also  at  the  time  the  first  be- 
comes due  ;  but  at  all  other  times  they  will  differ.  At  any 
time  between  the  day  of  their  date  and  the  day  when  the  first 
note  becomes  due,  the  true  value  of  the  second  note  will  bft 
greater  than  that  of  the  first ;  but  at  any  time  after  the  first 


298  BUSINESS    METHOD    OF    DISCOUNT. 

note  becomes  due,  the  true  value  of  the  first  note  will  b« 
greater  than  that  of  the  second.     Show  why  this  is  so. 

197.    Business  Method  of  Discount, 

(a.)  Business  men  are  usually  wiUing  to  allow  on  money 
paid  for  goods  before  it  is  due,  a  discount  equal  to,  or  greater 
than,  its  interest  from  the  time  of  payment  to  the  time  when, 
by  the  conditions  of  the  sale,  it  would  have  become  due. 

Thus,  when  only  the  interest  of  the  debt  is  discounted,  $824  due  in 
6  months,  interest  being  6  per  cent  per  year,  is  regarded  as  worth  $824 
—  .03  of  $824  =  $824  —  $24.72  =  $799.28,  whereas  it  ought  to  be  worth 

$800. 

(h.)  This  is  always  an  advantage  to  the  person  owing  the  money,  as 
it  enables  him  to  pay  his  debt  for  less  than  the  sum  which,  put  at  inter- 
est, would  amount  to  the  debt  at  the  time  it  would  become  due. 

(c.)  It  is  common  to  deduct  "as  much  as  five  per  cent  from  the  face 
of  a  bill  due  in  four  or  six  months,  and  even  more  is  sometimes  de- 
ducted. 

{d.)    The  present  worth  thus  obtained  may  be  called  the 

ESTIMATED  PRESENT  WORTH,  tO  distinguish  it  from  the  TRUE 

PRESENT  WORTH,  obtained  by  the  method  explained  in  106. 

1.  I  owed  a  debt  of  $8692,  payable  June  1,  1852 ;  but  my 
creditor,  offering  to  allow  me  discount  estimated  by  the  busi- 
ness method  at  the  rate  of  6  per  cent  per  year,  if  I  would 
pay  the  debt  Jan.  1,  1852, 1  borrowed  money  for  the  purpose 
at  6  per  cent  interest.  June  1,  1852,  I  paid  the  amount  of 
the  borrowed  money.  What  was  my  gain  by  the  transac- 
tion ? 

2.  Owing  a  debt  of  $1545,  due  in  6  months,  when  money 
is  worth  6  per  cent  per  year,  what  shall  I  gain  by  hiring 
money  enough  to  pay  it  now,  allowing  the  usual  business  dis- 
count on  the  debt,  and  then  paying  the  borrowed  money  with 
interest,  when  the  original  debt  would  otherwise  have  become 
due? 

8.   At  6  per  cent  per  year,  what  is  the  difference  between 


BUSINESS    METHOD    OP   DISCOUlfT.  299 

the  bank  discount  and  the  true  discount  of  a  note  for  $2059.40, 
payable  in  60  days  ?  * 

4.  Received  for  my  note  of  $600,  payable  in  6  months,  its 
true  present  worth.  How  much  more  did  I  receive  on  it  than 
I  should  have  received  at  a  bank,  money  being  worth  6  per 
cent  ?  How  much  interest  money  shall  I  have  gained,  when 
the  note  becomes  due,  over  what  I  should  have  gained  on  the 
present  worth,  as  determined  at  the  bank  ? 

5.  For  how  much  must  a  note,  payable  in  30  days,  be 
given,  that,  when  discounted  at  a  bank,  $900  may  bo  received 
on  it,  money  being  6  per  cent  ? 

Solution.  —  The  money  received  on  a  note  discounted  at  a  bank 
equals  the  sum  for  which  the  note  is  given,  minus  its  interest  for  the 
time  before  it  becomes  due.  Since,  at  6  per  cent,  the  interest  for  30 
days  and  grace,  or  33  days,  =  -^ff tr  =  2"^u^u  of  the  principal,  the  sum 
received  must  equal  f  B^^  —  j^^iy  =  2"§l^  of  the  face  of  the  note. 
Therefore  the  face  of  the  note  =  f^^gg-  of  the  sum  received  on  it, 
=  x^l^^  of  $900  =  $904,977,  or,  as  it  would  in  practice  be  considered, 
$904.98. 

The  numerical  work  can  be  expressed  thus  :  — 

33     _     11  2000 11_  _  1989        2000  ^  200000  __ 

6000  ~"  2000'       2000         2000  "~  2000*       1989  ^  ~~      221 

$904,977  ; 

hence  face  of  note  =  $904.98. 

6.  How  much  would  be  received  at  a  bank  on  a  note  for 
$904.98,  payable  in  30  days  ? 

Note.  —  The  above  example  suggests  the  method  of  proving  the  5th. 

7.  For  how  much  must  a  note  payable  in  3  months  be 
given,  that,  when  discounted  at  a  bank,  $1000  may  be  re- 
ceived on  it,  money  being  worth  6  per  cent  per  year  ? 

8.  For  how  much  must  a  note  payable  in  6  mo.  be  given, 
that,  when  discounted  at  a  bank,  $1800  may  be  received  on 
it,  money  being  worth  6  per  cent  per  year  ? 

*  Remember  that  three  days'  grace  are  always  allowed  on  a  note,  unless 
the  contrary  id  specified. 


500  TO    FIND    THE    RATE. 

9.  Obtained  at  a  bank,  on  my  note  payable  in  6  mo.,  money 
enough  to  buy  20  acres  of  land  at  $100  per  »vv.re.  The  day 
my  note  at  the  bank  became  due,  I  sold  the  land  for  $206:: .c)2 
cash.  Did  I  gain  or  lose  by  the  transaction,  and  how  much, 
money  being  worth  6  per  cent  per  year  ? 

10.  Obtained  at  a  bank,  on  my  note  payable  in  4  months, 
money  enough  to  buy  20  acres  of  land  at  $100  per  acre. 
The  day  the  note  became  due,  I  sold  the  land  for  casli,  at  such 
rate  that  the  price  of  18  acres  was  just  sufficient  to  pay  the 
note.  How  much  did  I  gain  by  the  transaction,  money  being 
worth  6  per  cent  ? 

11.  Bought  goods  to  the  amount  of  $864.27  on  a  credit  of 
6  months ;  but  the  seller  offering  to  deduct  5  per  cent  from 
the  face  of  the  bill  if  I  would  pay  cash,  I  hired  the  requisite 
amount  of  money,  giving  my  note  payable  in  6  months,  with 
interest  at  6  per  cent  per  year,  to  be  reckoned  from  date. 
For  how  much  less  than  the  value  of  the  original  bill  could  I 
pay  the  amount  of  this  note  ? 

12.  I  owed  $800,  due  in  6  months  ;  but  my  creditor  offer- 
ing to  deduct  5  per  cent  of  the  debt  for  cash,  I  paid  $380 
down.     How  much  did  I  still  owe  ? 

Suggestion.  —  Since  5  per  cent  of  the  debt  was  to  be  deducted  for 
cash,  the  cash  payment  would  be  95  per  cent,  =  -^^  =  ^o^,  of  the  part 
of  debt  it  would  cancel ;  or  the  part  cancelled  would  be  f^  of  the  cash 
paid. 

13.  I  owed  $900,  payable  in  4  months ;  but  my  creditor 
offering  to  deduct  4  per  cent  of  the  debt  for  ready  money,  I 
paid  $696  down.     How  much  did  I  still  owe  "^ 


198.    To  find  the  Rate. 

Problems  in  which,  the  principal,  interest,  and  time  being 
given,  we  are  required  to  find  the  rate,  rarely  occur  in  busi- 
ness life.  The  following  solutions  illustrate  the  principles 
which  apply  to  them. 


TO    FIND    THE    PRINCIPAL.  801 

1.  At  what  rate  per  cent  must  $648  be  on  interest  to  gain 
$81,873  in  2  yr.  3  mo.  17  da.  ? 

Solution.  —  The  principal  being  equal  to  648,000  mills,  and  the  inter- 
est to  81,873  mills,  the  interest  is  ^VfViht  =  T^inny  of  the  principal. 
2  yr.  3  mo.  17  da.  =  27  mo.  17  da.  =  827  da.  If  the  interest  for  827 
da.  =  T^%Vxy  of  the  principal,  the  interest  for  1  day  must  equal  ^^y 
of  Y^wQU  of  the  principal,  and  the  interest  for  1  yr.,  or  360  da.,  must 
equal  360  times  the  last  result,  or  ff^  of  j^xyJu  =  stjV  =  -05^  =  5^ 
per  cent. 

Required  the  rate  of  interest  when  — 

2.  $624  gains  $74.88  in  1  yr.  2  mo.  12  da. 

3.  $57.25  gains  $5,038  in  1  yr.  5  mo.  18  da. 

4.  $855  gains  $46.55  in  2  yr.  2  mo.  4  da. 

5.  $64.80  gains  $6,246  in  11  mo.  17  da. 


109.     To  jind  the  Principal  from  the  Interest. 

Problems  in  which,  the  interest,  rate,  and  time  being  given, 
we  are  required  to  find  the  principal,  are,  like  those  in  the 
last  article,  of  rare  occurrence. 

1.  What  principal  on  interest  at  6  per  cent  will  gain 
$37.47  in  1  yr.  3  mo.  ? 

Solution.  —  At  6  per  cent  per  year,  the  interest  of  any  principal  for 
15  mo.  =  jVff  =  -^-Q-  of  the  principal.  If  $37.47  is  ^  of  the  princi- 
pal, ^V  of  the  principal  must  equal  -j  of  §37.47,  and  the  principal  must 
equal  40  times  the  last  result,  or  ^-  of  $37.47,  which  is  $449.60. 

2.  What  principal  on  interest  at  8  per  cent  will  gain 
$26.18  in  1  yr.  4  mo.  15  da.  ? 

Solution.  —  Since  1  yr.  4  mo.  15  da.  =  le^-  mo.  =  1^7^-  =  "V^  of 
1  yr.,  the  interest  must  equal  -igl  of  8  per  cent,  or  11  per  cent  of  the 
principal.  If  $26.18  is  11  per  cent  of  the  principal,  1  per  cent  of  the 
principal  must  be  -yt  of  $26.18,  which  is  $2.38,  and  100  per  cent,  or  the 
principal,  must  equal  100  times  the  last  result,  which  is  $238. 

What  principal  on  interest  — 

3.  At  6  per  cent  will  gain  $8.73  in  5  mo.  ? 

4.  At  6  per  cent  will  gain  $4.77  in  1  yr.  5  mo.  20  da.  ? 

26 


302  COMPOUND    INTEREST. 

5.  At  5  per  cent  will  gain  $4.27  in  2  yr.  6  mo.  ? 

6.  At  7^  per  cent  will  gain  $116,127  in  4  yr.  4  mo.  4  da.  f 


dOO*     Compound  Interest. 

When  interest  is  to  be  paid  at  regular  intervals,  or,  if  nn- 
paid,  is  to  be  added  to  the  principal,  to  form  a  new  principal 
on  which  interest  is  to  be  computed,  it  is  called  Compound 
Interest.     The  following  example  illustrates  this  :  — 

1.  What  is  the  compound  interest  of  $784  for  2  yr.  8  mo. 
at  6  per  cent,  payable  annually  ? 

Solution. 
a  =  $784.       =  principal. 
06  of  a  =  b  =      47.04    =  interest  for  1st  year. 

a  -f-  b  =  c  =    831.04    =  amount  due  at  end  of  Ist  year. 

.06  of  c  =  d  =      49.862  =  interest  for  2d  year. 

c  +  d  =  e  =    880.902  =  amount  due  at  end  of  2d  year. 
.04  of  e  =  f  =      35.236  =  interest  for  8  mo. 


e  -|-  f  =  g  =    916.138  =  amount  due  at  end  of  2  yr.  8  mo. 
a  =    784.        =  principal. 

g  —  a  =  h  =  $132,138  =  compound  interest  for  2  yr.  8  mo. 

2.   To  what  sum  will  $437  amount  in  2  yr.  6  mo.  at  4  poi 
tent,  payable  semiannually  ? 

Solution. 
a  =  $437         =  principal. 
.02  of  a  =  b  =        8.74    =  interest  Ist  6  mo. 


a  +  b  =  c  =  445.74    =  amount  at  end  of  6  mo. 

.02  of  c  =  d  =  8.915  =  interest  2d  6  mo. 

c  +  d  =  e  =  454.655  =  amount  at  end  of  12  mo. 

.02  of  e  =  f  =  9.093  =  interest  3d  6  mo. 


e  -|-  f  =  g  =    463.748  =  amount  at  end  of  18  mo. 
02  of  g  =  h  =        9.275  =  interest  4th  6  mo. 

^  «|_  h  =  :  =    473.023  =  amount  at  end  of  2  yr 
.  "f  i  =  j  >=        G.i5^  ■«=  uiterest  5th  6  mo. 

'f.  j  -■»  i  *a  .,  ::?4JL483  ■=  ou»», ....»,  tfue  at  ond  ot  2  vr  $ 


TABLE    FOR    COMPOUND    INTEREST. 


308 


3.  What  is  the  compound  interest  of  $938.63  for  4  yr.  6 
mo.  at  6  per  cent,  payable  annually  ? 

4.  What  is  the  compound  interest  of  $573.32  for  2  yr.  3 
mo.  at  8  per  cent,  payable  quarterly  ? 

5.  To  what  sum  will  $1000  amount  in  3  yr.  2  mo.  at   6 
per  cent,  payable  annually  ? 

6.  To  what  sum  will  $500  amount  in  4  yr.  3  mo.  at  5  per 
cent,  payable  semiannually  ? 

SOI.    TaUe  for   Gompound  Interest. 

Table  showing  what  part  of  any  principal  is  equal  to  its 
amount  at  3,  4,  5,  6,  7,  and  8  per  cent  annual  compound 
interest,  for  any  number  of  years  not  exceeding  25. 


Yrs. 
1. 

3  per  cent. 

4  per  cent. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

8  per  cent. 

1.03 

1.04 

1.05 

1.06 

1.07 

1.08 

2. 

1.0609 

1.0816 

1.1025 

1.1236 

1.1449 

1.1664 

3. 

1.092727 

1.124864 

1.157625 

1.191016 

1.225043 

1.259712 

4. 

1.125509 

1.169858 

1.215506 

1.262477 

1.310796 

1.360489 

5. 

1.159274 

1.216653 

1.276281 

1.338225 

1.402552 

1.469328 

6. 

1.194052 

1.265319 

1.340096 

1.418519 

1.500730 

1.586874 

7. 

1.229874 

1.315932 

1.407100 

1.503630 

1.605781 

1.713824 

8. 

1.266770 

1.368569 

1.477455 

1.593848 

1.718186 

1.850930 

9. 

1.304773 

1.423312 

1.551328 

1.689479 

1.838459 

1.999005 

10. 

1.343916 

1.480244 

1.628894 

1.790848 

1.967151 

2.158925 

11. 

1.384234 

1.539454 

1.710339 

1.898298 

2.104852 

2.331639 

12. 

1.425761 

1.601032 

1.795856 

2.012196 

2.252191 

2.518170 

13. 

1.468534 

1.665073 

1.885649 

2.132928 

2.409845 

2.719624 

14. 

1.512590 

1.731676 

1.979931 

2.260903 

2.578534 

2.937194 

15. 

1.557967 

1.800943 

2.078928 

2.396558 

2.759031 

3.172169 

16. 

1.604706 

1.872981 

2.182874 

2.540351 

2.952164 

3.425943 

17. 

1.652848 

1.947900 

2.292018 

2.692772 

3.158815 

3.700018 

18. 

1.702433 

2.025816 

2.406619 

2.854339 

3.379932 

3.996019 

19. 

1.753506 

2.106849 

2.526950 

3.025599 

3.616527 

4.315701 

20. 

1.806111 

2.191123 

2.653298 

3.207135 

3.869684 

4.660957 

21. 

1  860294 

2.278768 

2.785962 

3.399563 

4.140562 

5.033834 

22. 

1.916103 

2.369918 

2.925260 

3.603537 

4.430402 

5.436540 

23. 

1.973586 

2.464715 

3.071524 

3.819749 

4.740530 

5.871464 

24. 

2.032794 

2.563304 

3.225100 

4.048934 

5.072367 

6.341181 

25. 

2.093778 

2.665836 

3.386355 

4.291870 

5.427433 

6.848475 

B04  ,  COMMISSION. 

Use  of  the  Table.  —  The  amouDt  of  any  sum  of  money, 
at  compound  annual  interest,  for  any  time  and  rate  mentioned 
in  the  table,  may  be  found  by  multiplying  the  principal  by  the 
appropriate  number  selected  from  the  table.  This  will  be 
illustrated  in  the  following  examples  and  solution. 

1.  What  is  the  amount  of  $245.73  for  12  yr.  6  mo.  20  da. 
annual  compound  interest  at  6  per  cent  ? 

Solution.  —  By  the  table,  it  appears  that  the  amount  of  a  sum  for 
12  years,  at  6  per  cent,  compound  interest,  is  2.012196  times  the  princi- 
pal. Multiplying  $245.73  by  2.012196,  we  have  $494.45692308,  or,  omit- 
ting the  denominations  below  mills, 

a  =  $494,457  =  amount  for  12  years. 
^  of  a  =  b  =      16.481  =  interest  for  6  mo.  20  da. 


a  +  b  =  c  =  $510,938  =  amount  for  12  yr.  6  mo.  20  da 

What  is  the  amount  at  annual  compound  interest  of — 

2.  $578.67  for  11  yr.  4  mo.  at  3  per  cent  ? 

3.  $147.43  for  22  yr.  4  mo.  24  da.  at  5  per  cent? 

4.  $1467  for  18  yr.  at  7  per  cent  ? 
What  is  the  compound  interest  of  — 

5.  $1137.38  for  13  yr.  6  mo.  at  6  per  cent  ? 

6.  $328.96  for  9  yr.  3  mo.  at  4  per  cent  ? 


303.    Commission, 

Money  received  for  services  in  buying  and  selling  goods 
for  others  is  called  Commission,  and  is  usually  reckoned  at  a 
certain  per  cent  of  the  cost  of  the  goods  bought,  or  price  of 
hose  sold.  A  merchant  who  makes  it  his  business  to  buy 
and  sell  on  commission  is  called  a  commission  merchant. 
(See  177th  page,  Ex.  21,  Note.) 

1.  A  commission  merchant  sold  goods  for  a  manufacturer 
to  the  amount  of  $568.36,  for  which  he  charged  a  commission 
of  2^  per  cent.  What  was  his  commission,  and  how  much 
will  be  due  to  the  manufacturer  ? 


COMMISSION.  805 

Ansrm.  — His  commission  =  .02|  of  $568.36  =  $li.209,  and  the 
snra  due  the  manufacturer  =  $568.36  —  $14.21  =  $554.15. 

2.  A  commission  merchant  bought  goods  for  a  country 
trader  to  the  amount  of  $738.27,  for  which  he  charged  a  com- 
mission of  1^  per  cent.  What  was  his  commission,  and  what 
sum  must  the  trader  remit  to  pay  for  the  goods  and  commis- 
sion? 

Answer.  —  His  commission  =  .Ol^-  of  $738.27  =  $11.07,  and  the 
trader  must  remit  $738.27  +  $11.07  =  $749.34. 

3.  I  sold  for  Seth  Jones  2024  pounds  of  butter  at  1 9  cents 
per  pound,  and  5276  pounds  of  cheese  at  1^  cents  per  pound. 
What  was  the  vahie  of  my  commission  at  2 J-  per  cent  on  the 
sales  ?     How  much  money  ought  I  to  pay  him  ? 

4.  I  bought  for  Francis  Jackson  50  pairs  of  boots  at 
$3.87 J  per  pair,  100  pairs  at  $2.75  per  pair,  75  pairs  at 
$2.1 6|  per  pair,  100  pairs  of  shoes  at  $1.56  per  pair,  and  80 
pairs  at  $1.08  per  pair.  What  was  the  value  of  my  commis- 
sion at  2  per  cent  on  the  purchase  ?  How  much  money  must 
Mr.  Jackson  remit  to  me  to  pay  for  the  goods  and  my  com- 
mission ? 

5.  1  have  sent  $5000  to  my  agent  in  New  Orleans,  direct- 
ing him  to  expend  it  for  cotton,  first  deducting  his  commission 
of  2  per  cent  on  the  purchase.  What  wiU  be  his  commission, 
and  what  will  he  expend  for  cotton  ? 

Suggestion.  —  The  $5000  sent  includes  the  sura  to  be  actually  invested 
and  my  agent's  commission  of  2  per  cent  on  that  sum,  and  is  therefore 
\%%  of  the  purchase  money.  Hence,  the  purchase  money,  or  sum  to  be 
expended  by  my  agent,  is  \%^  of  $5000,  and  his  commission  is  y§3  of 
$5000.     For  proof,  see  if  the  sum  expended,  plus  .02  of  it,  equals  $5000. 

6.  I  have  received  $5600  from  my  correspondent  in  St. 
Louis,  with  directions  to  expend  it  in  merchandise,  first*  de- 
ducting my  commission  of  2|-  per  cent  on  the  money  expend- 
ed.    How  much  ought  I  to  expend  ? 

7.  INIy  agent  in  Rochester  writes  that  he  has  purchased  a 
lot  of  flour  for  me  at  $5  per  barrel,  and  that  the  entire  cost 

26* 


306  COMMISSION. 

of  the  flour,  including  his  commission  of  2^  per  cent,  is  $1 600 
How  many  barrels  of  flour  were  purchased  ? 

8.  Briggs  «fc  Grant  sold  for  Jenks,  Clarke,  &  Co.  1000 
brooms  at  $.25  apiece,  for  which  they  charge  a  commission  of 
4  per  cent.  Pursuant  to  instructions,  they  invest  the  balance 
in  sugar  at  8  cents  per  pound,  first  deducting  their  commission 
of  2  per  cent  on  the  purchase.  How  many  pounds  of  sugar 
did  they  buy  ? 

9.  My  agent  at  Cincinnati  writes  that  he  has  purchased  on 
my  account  a  lot  of  provisions,  and  that  his  commission  of  1^ 
per  cent  on  the  purchase  is  $13.50.  How  many  dollars 
worth  of  provisions  has  he  bought  ? 

10.  My  agent  at  New  Orleans  has  purchased  for  me  a  lot 
of  cotton  at  6  cents  per  pound,  for  which  he  charges  a  com- 
mission of  1§  per  cent.  His  commission  amounts  to  $38.80. 
How  many  pounds  of  cotton  has  he  bought,  and  how  much 
money  must  I  remit  to  him  to  pay  for  the  cotton  and  his  com- 
mission ? 

11.  Haskell  &  Latham  sell  for  me,  at  auction,  goods  to  the 
amount  of  $8732.  Their  charges  are  as  follows :  Auction 
tax,  1  per  cent ;  commission,  4  per  cent ;  guarantying  the 
sales,  3  per  cent ;  advertising,  $23.25  ;  truckage  and  storage, 
$11.25.     How  much  money  will  be  due  me  ? 

Suggestion.  —  The  auction  tax,  commission,  and  guaranty  amount  to 
8  per  cent  of  the  value  of  the  goods,  which,  together  with  the  other  ex- 
penses, must  be  deducted  from  the  value  of  tiie  goods,  to  leave  the  sum 
due  me. 

12.  Lewis  &  Johnson  sell  for  Field  &  Dean  96  cases  of 
cassimere,  each  case  containing  276  yards,  at  $1.25  per  yard, 
for  which  they  charge  a  commission  of  3^  per  cent.  They 
receive  instructions  from  Field  &  Dean  to  remit  them  half  of 
the  net  proceeds,  and  to  invest  the  remainder,  after  deducting 
a  commission  of  1^  per  cent  on  the  purchase,  in  wool,  at  50 
cents  per  pound.  What  was  the  value  of  the  cash  remitted 
to  Field  &  Dcim  ?     How  many  pounds  of  wool  were  bought  ? 


STOCKS.  807 

203.    Stocks, 

(a.)  Money  invested  in  any  property  designed  to  yield  an 
income  is  called  stock. 

The  money  invested  by  a  man  in  his  business  is  called  his  stock  in 
TRADE ;   that  invested  in   government  securities,  bonds,  &c.,  is   called 

GOVEHN3IENT   STOCK. 

(5.)  The  CAPITAL  STOCK  of  any  incorporated  company  is 
the  money  paid  in  by  its  members  for  the  general  purposes 
for  which  the  company  was  formed.  It  is  divided  into  equal 
parts,  called  shares.  Any  person  owning  one  or  more  of 
these  shares  is  a  stockholder,  or  member  of  the  corpora- 
tion. Stocks  is  a  general  term  applied  to  the  shares  them- 
selves, which  may  be  bought  or  sold  like  any  other  property. 

(c.)  The  value  at  which  the  shares  of  any  corporation  are 
rated  in  estimating  its  capital  stock,  that  is,  their  first  or  origi- 
nal value,  is  called  their  nominal  value,  or  their  par  value, 
and  is  always  tlie  same.  The  price  which  they  will  bring,  if 
exposed  for  sale,  is  their  true  or  real  value,  and  is  differ- 
ent at  different  times.  If  the  real  value  equals  the  par  value, 
the  stocks  are  at  par ;  if  it  be  greater,  they  are  above  par, 
and  sell  at  a  premium,  or  advance  ;  if  it  be  less,  they  are 
below  par,  and  sell  at  a  discount. 

(d.)  Tlie  profits  accruing  to  the  corporation,  if  any,  are  at 
intervals  distributed  among  the  members  in  proportion  to  the 
number  of  shares  each  holds,  and  are  then  called  dividends. 
The  dividends  are  usually  reckoned  at  a  certiiin  per  cent  of 
the  par  value  of  the  shares. 

1.  How  much  will  11  shares  Fall  River  Railroad  stock 
cost,  at  an  advance  of  6  per  cent,  the  par  value  being  $100 
per  share  ? 

Solution. 
$1100  =  par  value  of  11  shares. 
66  =  6  per  cent  premium. 

$1166  =  real  value,  or  required  cost. 

2.  How  much  will  11  shares  Providence  Railroad  stock 


608  STOCKS. 

cost,  at  a  discount  of  6  per  cent,  the  par  value  being  $100 
per  share  ? 

Solution. 
$1100  =  par  value  of  11  shares. 
66  =  6  per  cent  discount. 


$1034  =  real  value,  or  required  cost. 

3.  How  much  will  53  shares  Suffolk  Bank  stock  cost  at  an 
idvance  of  23  per  cent,  the  per  value  being  $100  per  share  ? 

4.  How  much  will  43  shares  Vermont  Central  Railroad 
stock  cost,  at  a  discount  of  78  per  cent,  the  par  value  being 
^50  per  share  ? 

5.  How  many  shares  of  stock  at  an  advance  of  5  per  cent 
on  the  par  value  of  $100  per  share,  can  be  bought  for  $1995  ? 

6.  How  many  shares  of  stock  at  a  discount  of  5  per  cent 
from  the  par  value  of  $100  per  share,  can  be  bought  for 
$1805  ? 

7.  A  broker  paid  $1776.50  for  stock  at  an  advance  of  4^ 
per  cent.     What  was  the  par  value  of  the  stock  bought  ? 

8.  A  broker  paid  $4850  for  bank  stock,  at  a  discount  of  3 
per  cent.     What  was  the  nominal  value  of  the  stock  bought  ? 

9.  A  broker  sold  on  consignment  376  shares  bank  stock, 
par  value  $100  per  share,  at  an  advance  of  7  per  cent.  He 
charged  25  cents  per  share  for  his  services.  How  much 
ought  he  to  remit  to  the  person  consigning  the  stock  ? 

10.  I  bought  40  shares  of  railroad  stock,  at  33  per  cent  b&- 
low  par,  and  after  keeping  them  10  months,  sold  them  at  20 
per  cent  below.  How  much  did  I  gain  on  them,  allowing 
that  I  hired  money  for  the  investment  at  6  per  cent  interest, 
and  that  the  par  value  of  the  shares  was  $100  each  ? 

11.  Alfred  E.  Potter  bought  10  shares  of  bank  stock  at 
par,  which  was  $50  per  share.  At  the  end  of  3  months  he 
received  a  dividend  of  4  per  cent,  and  at  the  end  of  9  months 
he  received  another  of  3  per  cent.  At  the  end  of  one  year 
he  sold  the  stock  at  an  advance  of  3  per  cent.  Money  being 
worth  6  per  cent  per  year,  how  much  did  he  gain  by  the 
transactions  ? 


TXSURAXCE.  309 

NoTP  —  Interest  should  be  reckoned  on  the  dividends  from  the  time 
Jiey  were  made. 

.  12.  Crocker  &  Guild  sent  $972.63  to  a  stock  broker, 
directing  him  to  invest  it  in  Fall  River  Railroad  stock.  He 
bought  the  stock  at  a  premium  of  7  per  cent,  the  par  value 
being  $100  per  share,  and  he  charged  a  commission  of  1  per 
cent  on  the  money  invested.     How  many  shares  did  he  buy  ? 

13.  I  paid  $7398  for  stock  at  10  per  cent  below  par,  and 
some  time  after  sold  the  stock  at  10  per  cent  above  par.  How 
much  did  I  gain  on  it  ? 

14.  Mr.  Hamblin  bought  stock  at  10  per  cent  above  par, 
but  he  was  obliged  to  sell  it  at  10  per  cent  below  par.  Allow 
ing  that  he  lost  $138.40  on  it,  what  did  he  pay  for  ife.? 

9041.    Insurance. 

(a.)  Instjrance  is  an  obligation  assumed  by  one  individv  rf 
or  company  to  pay  to  another  a  certain  sum  of  money  on  t).e 
occurrence  of  some  contingent  event. 

(6.)  A  house  is  insured  against  loss  by  fire,  when  some  individual  or 
company  agrees  to  pay  the  owner  a  specified  sura  if  it  is  burned  within  a 
given  time. 

(c.)  A  ship  in  like  manner  may  be  insui-ed  against  loss  by  fire  or  by 
any  of  the  perils  of  the  sea. 

[d.)  When  the  obligation  is  to  pay  the  person  insured  a  certain  sura 
if  he  is  sick,  it  is  called  health  insurance  ;  when  it  is  to  pay  to  his 
heirs,  or  to  some  particular  person,  a  specified  sum  if  the  person  insured 
dies,  it  is  called  life  insurance. 

(e.)    The  written  certificate  of  insurance  is  called  a  policy. 

{f.)  The  sum  paid  for  insurance  is  called  the  premium. 

{g.)  If  property  is  insured,  the  premium  is  a  certain  per  cent  of  the 
8um  covered,  i.  e.,  of  the  sum  for  which  it  is  insured,  and  is  usually  paid 
at  the  time  of  effecting  the  insurance. 

[h.)  When  health  or  life  is  insured,  the  premium  is  usually  a  given 
sum  paid  annually  during  the  time  for  which  the  insurance  continues ; 
and  its  amount  is  calculated  from  tables  prepared  by  the  different  insur- 
ance companies.  It  will,  therefore,  be  unnecessary  to  give  examples  in 
life  or  health  insurance. 

1.   What  would  be  the  expense  of  getting  $1000  insured 


510  ASSESSMENT    OP   TAXES. 

on  a  house  at  a  premium  of  1^  per  cent,  the  charge  for  the 
policy  being  $1  ? 

2.  I  had  $1000  insured  on  my  house  for  7  years,  for  which 
I  paid  a  premium  of  2^  per  cent,  and  $1  for  the  policy.  At 
the  end  of  5  yrs.  7  mo.  15  da.  the  house  was  burned,  and  I 
received  the  amount  for  which  it  was  insured.  Money  being 
worth  6  per  cent  per  year  compound  interest,  how  much 
did  I  really  save  by  having  the  house  insured  ? 

3.  Oct.  1,  1854,  I  bought  a  lot  of  flour  for  $6000,  giving 
my  note  payable  in  6  months,  and  immediately  shipped  it  for 
England.  Oct.  8,  1854, 1  got  it  insured  for  $6000,  paying  a 
cash  premium  of  1^  per  cent,  and  $1  for  the  policy.  Oct.  10, 
1854,  I  ^aid  a  bill  of  $50  for  cartage  and  other  expenses. 
Nov.  1,  1854,  I  received  information  that  the  vessel  was  lost 
at  sea ;  and  Dec.  1,  the  insurance  company  paid  me  the  in- 
surance. I  immediately  put  it  on  interest  at  6  per  cert,  till 
my  note  for  the  flour  became  due,  when  I  made  a  complete 
settlement.  Had  I  gained  or  lost  by  the  transactions,  and 
how  much  ? 

205,    Assessment  of  Taxes. 

(a.)  Taxes  are  duties  or  assessments  laid  on  persons  or 
property,  usually  for  some  public  purpose. 

(5.)  A  tax  on  persons  is  called  a  poll  tax,  and  a  tax  on 
property  is  called  a  property  tax. 

The  poll  tax  is  only  assessed  on  males  between  certain  ages,  as  be- 
tween twenty-one  and  seventy ;  and  in  some  states  is  not  assessed 

(c.)    In  assessing  taxes,  it  is  necessary,  — 

First.  To  estimate  the  value  of  all  the  property  to  be 
taxed,  and  make  a  complete  inventory  of  it. 

Second.  To  find  the  number  of  polls,  i.  e.,  the  number  of 
persons  liable  to  pay  a  poll  tax. 

Third.  To  determine  what  portion  of  the  tax  is  to  be 
raised  upon  the  polls,  and  to  divide  it  equally  among  them. 

Fourth.  To  find  how  much  must  be  paid  on  each  dollar  of 
the  taxable  property,  to  raise  the  remainder  of  the  tax. 


ORDERS.  811 

This  last  may  be  done  by  dividing  the  amount  to  be  raised  by  the 
estimated  value  of  the  property  on  which  it  is  to  be  raised.  It  will  then 
be  easy  to  find  the  tax  of  any  individual. 

1.  A  tax  of  $4800  is  to  be  raised  by  a  certain  town.  The 
taxable  property  is  valued  at  $960,000,  and  there  are  320 
polls,  each  taxed  $1.50.  What  will  be  the  tax  on  each  dol- 
lar, and  what  will  be  the  tax  of  each  of  the  follow^ing  persons  ? 

A,  who  pays  a  tax  on  $5700,  and  2  polls. 

B,  who  pays  a  tax  on  $728,  and  1  poll. 

C,  who  pays  a  tax  on  $8976,  and  3  polls. 

D,  who  pays  a  tax  on  $1147,  and  1  poU. 

Solution.  —  The  tax  on  320  polls  at  $1.50  each  is  $480,  which,  sub- 
tracted from  $4800,  leaves  $4320  to  be  assessed  on  the  property. 

Since  960,000  is  to  be  taxed  $4320,  one  dollar  will  be  taxed  g^6o\>07 
ef  $4320,  which  is  A^  mills. 

A's  tax  on  2  polls  would  be  twice  $1.50,  or  $3,  and  on  $5700  prop- 
erty would  be  5700. times  A^  mills,  which  is  $25.65,  and  added  to  $3 
gives  $28.65  as  the  amount  of  A's  tax. 

The  tax  of  the  others  may  be  found  in  the  same  way. 

2.  A  tax  of  $4800  is  to  be  raised  in  a  certain  town.  The 
taxable  property  is  valued  at  $1,228,000,  and  there  are  575 
polls  to  be  taxed  $1.40  each.  How  much  is  the  tax  on  $1  ? 
How  much  is  the  tax  on  each  of  the  following  persons  ? 

J.  S.,  who  pays  a  tax  on  $1728,  and  1  poll. 
S.  E..,  who  pays  a  tax  on  $4235,  and  2  polls. 
L.  M.,  who  pays  a  tax  on  $7945,  and  no  poll. 
F.  Gr.,  who  pays  a  tax  on  $2794,  and  1  poll. 


a06.     Orders. 

(a.)  If  a  person  should  wish  to  have  money  which-  is  due 
him  paid  to  some  one  else,  he  might  write  an  order  for  it, 
i.  e.,  a  paper  requesting  the  one  who  owes  it  to  pay  it  to  the 
third  person. 

(6.)  Suppose,  for  instance,  that  Edward  Mellen  owes  Richard  Jack- 
son five  hundred  dollars,  and  that  Mr.  Jackson  owes  Stephen  Baker 
one  huudred  dollars. 


312  ORDERS. 

(c.)  Suppose  that  Mr.  Baker  presents  his  bill  for  payment,  and  tha 
Mr.  Jackson,  not  having  the  money  by  him,  gives  him  the  following 
order  on  Mr.  Mellen  :  — 

Q>hma,%^    8tt)eEEen-,    ®*cjy. 

t)  teo-^e    kau,    to    lae    oiael    op    tjtckoerv    ^!6a<Ae\ 
one.    ruwvo-Veo-    aobla/ii',    cuvo-    cHaVae    trie    ^cuvve    to-    rrte.   ^ 

t%i/clvaio    J^acR^orv. 

(<f.)  In  considering  this  order,  we  may  notice,  — 

First.  The  "  $100"  and  the  date  at  its  head.      See  182,  (d.) 

Second.  "  Edward  Mellen,  Esq.,"  which  is  written  to  show  to  whom 
the  order  is  addressed. 

Third.  The  request,  "  Please  pay  to  the  order  of  Stephen  Baker  one 
hundred  dollars."  The  explanations  of  182,  {d.)  will  show  the  use  of 
the  various  parts  of  this. 

Fourth.  "  And  charge  the  same  to  me,"  which  authorizes  Mr.  Mellen 
to  charge  the  money  to  Mr.  Jackson,  as  though  it  had  been  paid  directly 
to  him.    The  phrase  "  on  my  account "  would  mean  the  same  thing. 

(e.)  This  order,  when  written,  would  be  taken  to  Mr.  Mellen  by  Mr. 
Baker.  If  Mr.  Mellen  pays  it,  Mr.  Baker  indorses  it,  or  writes  the 
words  "  Received  payment,"  with  his  name,  upon  it,  and  hands  it  to  Mr. 
Mellen,  who  keeps  it  as  evidence  that  he  has  actually  paid  out  so  much 
money  on  account  of  Mr.  Jackson. 

(/)  By  this  arrangement  it  will  be  seen  that  Mr.  Jackson  pays  tho 
one  hundred  dollars  which  he  owes  to  Mr.  Baker,  and  that  Mr.  Mellen 
pays  one  hundred  dollars  of  his  indebtedness  to  Mr.  Jackson. 

{g.)  If  Mr.  Mellen  should  refuse  to  pay  the  above  order,  Mr.  Baker 
would  have  no  right  to  commence  legal  proceedings  against  A/m,  but  he 
would  have  the  same  claim  against  Mr.  Jackson  that  he  had  before  the 
order  was  given. 

{h.)  If  Mr.  Mellen  is  willing  to  pay  it,  but  is  unable  to  do  so  the  day 
that  it  is  presented,  he  should  accept  it,  by  writing  his  name,  or  the  word 
♦'  accepted  "  and  his  name,  upon  it.  This  is  called  an  acceptance,  and 
would  give  Mr.  Baker  the  same  legal  claims  against  Mr.  Mellen  that  he 
would  have  had  upon  a  note  for  the  same  amount. 

(t .)   An  order  may  be  given  to  pay  in  a  given  time,  as  "  thirty  dayi 


BILLS    OF    EXCHANGE.  313 

after  date."  Such  an  order  should  at  once  be  presented  for  acceptance. 
80  that  the  person-  on  whom  it  is  drawn  may  be  held  responsible  fbr  its 
payment. 

{j.)  An  order  may  be  payable  in  some  given  time  after  sight,  i.  e.,  in 
some  given  time  after  it  is  exhibited  to  and  accepted  by  the  person  to 
whom  it  is  directed.  Usage  varies  with  regard  to  grace  on  such  orders, 
tliongh  it  is  commonly  allowed. 


307.    BiUs  of  Exchange. 

(a  )  "When  an  order  is  drawn  on  a  person  living  in  a  distant  place,  it 
is  called  a  bill  op  exchange. 

(6.)  Bills  of  exchange  are  of  two  kinds  —  foreign  and  inland;  the 
former  being  drawn  on  persons  living  in  foreign  countries,  and  the  latter 
on  those  living  within  our  own.  There  are,  however,  no  other  essential 
differences  between  foreign  and  inland  bills,  than  such  as  are  connected 
with  the  different  business  customs  existing  in  different  countries. 

(c.)  It  is  customary  to  write  two  or  three  copies  of  a  foreign  bill  of 
exchange,  and  to  send  them  by  different  vessels,  so  as  to  render  it 
reasonably  certain  that  one  of  them  shall  reach  the  person  to  whom  it  is 
sent.  These  bills  constitute  a  set  op  exchange,  and  are  called  the 
first,  second,  and  third  of  exchange.  They  are  so  worded  as  to  express 
that  the  payment  of  one  renders  the  others  void. 

{d.)   The  following  illustrates  their  form  :  — 

©occh/ctrvae    Poi    ok)  1000  • 

U  u>enkw    ocuti    ahe.'v    ilahl    op    Uiiiy    PLwt    op    esccilcuvoe, 
fiecorvd    cuvd   Outd  lui/W/CUO-^    kcut   to    tae    oidei    oi   ^M'iou>n,    cuv(^ 
tMuKek,    oive    wvou/JKLao    koiuvo^,    orvo    cK/O/iae    th,e    ^cwwe    to     ou%, 
accoiutfc.  xi^ottet    ^     O^amm-ojv^. 

e2^ui-eUtoot. 

(«.)  The  other  bills  of  the  set  would  be  dated  and  directed  like  t&iSj 
but  the  second  would  read,  — 

27 


314  BILLS    OF    EXCHANGE. 

"  Twenty  days  after  sight  of  this  second  of  exchange,  (first  and  thW 
unpaid,)  pay,"  &c.,  as  before. 

A  corresponding  change  would  be  made  in  the  third. 

{/.)  To  illustrate  the  use  of  such  biUs,  let  us  suppose  that  Brown 
and  Butler,  of  Boston,  owe  Miller  and  Jones,  of  Liverpool,  one  thon- 
gand  pounds,  and  that  French,  Harris,  &  Co.,  of  Liverpool,  owe  Potter 
and  Hammond,  of  Boston,  one  thousand  pounds. 

(g.)  It  is  obvious  that  if  the  parties  should  pay  their  debts  in  specie, 
they  must  incur  the  expense  and  risk  of  transporting  two  thousand 
pounds  across  the  Atlantic,  and  that  then  as  much  money  would  have 
been  brought  back  to  each  country  as  has  been  sent  out  from  it. 

(h.)  But  if  Brown  and  Butler  should  buy  of  Potter  and  Hammond, 
a  set  of  exchange  for  one  thousand  pounds  on  French,  Harris,  &  Co., 
and  should  indorse  it  to  Miller  and  Jones,  and  send  it  to  them,  and  they 
should  collect  it,  all  these  debts  would  be  cancelled  without  any  trans- 
fer of  specie  from  one  country  to  the  other. 

(i.)  For  Potter  and  Hammond  would  have  received  from  Brown  and 
Butler  an  equivalent  for  their  claim  against  French,  Harris,  &  Co.,  and 
have  thus  got  the  means  of  paying  Miller  and  Jones.  Miller  and  Jones 
would  have  received  from  French,  Harris,  &  Co.  an  equivalent  for  their 
claim  against  Brown  and  Butler ;  and  hence  each  would  have  paid  out 
the  value  of  what  he  owed,  and  have  received  the  value  of  what  was 
due  to  him. 

(j.)  Such  transactions  as  the  above  are  so  manifestly  to  the  advan- 
tage of  the  parties  concerned  in  them,  that  a  demand  for  bills  of  ex- 
change would  naturally  spring  up  in  countries  having  much  business 
intercourse  with  each  other,  as  between  the  United  States  and  England, 
or  the  United  States  and  France. 

(k.)  If  our  merchants  should  owe  the  merchants  of  England  more 
than  they  owe  us,  there  would  be  more  persons  here  wishing  to  buy  than 
sell  bills  on  England.  The  demand  would  thus  be  greater  than  the 
supply,  and  bills  would  bring  more  than  the  sum  fur  which  they  were 
drawn,  or  would  be  at  a  premium. 

{I.)  If,  however,  English  merchants  should  owe  us  more  than  we  owe 
them,  more  of  our  merchants  would  wish  to  sell  than  to  buy,  and  bills 
on  England  would  be  at  a  discount. 

(m.)  There  would  be  a  limit  to  the  premium  on  one  side,  and  to  the 
discount  on  the  other ;  for  merchants  would  not  pay  more  premium  for 
bills  of  exchange  than  it  would  cost  them  for  freight  and  insurance  to 
transport  specie  across  the  ocean. 

(n.)  When  exchange  on  a  foreign  country  is  at  a  premium,  it  is  said 
to  be  against  us,  for  it  not  only  makes  our  merchants  pay  more  than  the 
amount  of  their  debts,  but  it  indicates  that  a  balance  of  specie  is  due  to 


BILLS    OF    EXCHANGE.  ^  315 

that  country.  When  it  is  at  a  discount,  it  is  said  to  be  in  our  favor,  for 
it  not  only  enables  our  merchants  to  pay  their  debts  for  less  than  their 
amount,  but  indicates  that  a  balance  of  specie  is  due  to  us. 

(o.)    These  variations  in  value  are  called  the  course  of  exchange. 

(p.)  It  is  an  important  thing  with  the  merchant  to  watch  the  course 
of  exchange  between  different  countries,  as  he  can  often  save  consider- 
able money  by  making  his  remittances  tbrough  an  indirect  route.  Thus, 
it  may  happen  that  bills  on  England  command  a  much  higher  premium 
than  bills  on  France ;  at  the  same  time  it  may  happen  that  exchange 
from  France  on  England  can  be  bought  at  a  low  rate.  In  such  circum- 
stances it  may  happen  that  by  first  buying  a  bill  on  France,  and  then 
sending  it  to  a  banking  house  there,  with  directions  to  buy  a  bill  on 
England,  a  merchant  may  pay  his  debt  for  less  than  he  could  by  direct 
remittance. 

(q.)  The  pound  sterling  varies,  as  has  been  said,  from  $4.83  to  $4.86, 
but  in  exchange  its  par  value  is  assumed  to  be  $4.44|^ ;  so  that  wben  a 
pound  of  exchange  brings  its  true  value,  it  should  be  very  nearly  9  per 
cent  above  par. 

1.  Paine  &  Colman,  of  Providence,  bought  of  Robert 
Greene  &  Co.  a  set  of  exchange  for  £1500  on  R.  &  S.  Hub- 
bard, of  Liverpool,  paying  10  per  cent  premium.  How  many 
dollars  did  it  cost  them  ? 

2.  What  will  a  set  of  exchange  on  London  for  £2000  cost 
at  9f  per  cent  premium  ? 

3.  What  will  a  set  of  exchange  on  France  for  5000  francs 
cost  at  1  per  cent  discount,  the  par  value  of  the  franc  being 
$.186? 

4.  Which  will  be  the  most  profitable,  and  how  much  the 
most,  to  buy  a  set  of  exchange  on  London  for  £1500  af  10^ 
per  cent  premium,  or  to  buy  sovereigns  enough  at  $4.86  to 
pay  the  debt,  allowing  that  it  will  cost  2^  per  cent  of  their 
value  for  freight  and  insurance  ? 

5.  How  much  will  it  cost  to  remit  £3000  to  Liverpool 
through  France,  when  exchange  on  France  can  be  bought  at 
the  rate  of  5.4  francs  for  $1,  and  exchange  on  England  can 
be  bought  in  France  at  the  rate  of  £1  for  24  francs,  allowing 
that  a  banking  house  in  Paris  will  charge  ^  per  cent  for  pur- 
chasing the  exchange  in  that  city  ? 


316  '  I'llOFIT    AND    LOSS. 

308,    Profit  and  Loss. 

Most  problems  in  profit  and  loss  come  under  one  of  three 
classes,  viz. :  — 

First.  Those  in  which  it  is  required  to  find  for  what  price 
articles  must  be  sold  that  their  owner  may  gain  or  lose  a  cer* 
tain  per  cent  of  their  cost. 

Second.  Those  in  which  it  is  required  to  find  what  per 
cent  of  the  given  cost  will  be  gained  or  lost  by  selling  articles 
at  a  given  price. 

Third.  Those  in  which  it  is  required  to  find  the  cost,  or 
some  per  cent  of  the  cost,  of  articles  on  which  a  certain  per 
cent  will  be  gained  or  lost  by  selling  them  at  a  given  price. 

They  involve  similar  principles  to  those  involved  in  other 
examples  in  percentage  and  fractions. 

1.  A  speculator  bought  a  lot  of  land  for  $2473.  For  how 
much  must  he  sell  it  to  gain  25  per  cent  of  its  cost  ? 

Suggestion.  —  He  raust  sell  it  for  the  cost,  plus  25  per  cent  of  the 
cost ;  or,  since  25  per  cent  =  ^,  he  must  sell  it  for  the  cost,  plus  ^  of 
the  cost.* 

2.  Mr.  Huntington  bought  a  lot  of  grain  at  54  cents  per 
bushel ;  but  it  being  damaged,  he  was  obliged  to  sell  it  at  a 
loss  of  16f  per  cent.     For  how  much  per  bushel  did  he 

sell  it  ? 

Suggestion.  —  Since  16f  per  cent=  |-,  he  must  have  sold  it  for  the 
cost,  minus  ^  of  the  cost.* 

3.  Mr.  Shelley  bought  a  lot  of  sugar  at  8  cents  per  pound. 
For  how  much  per  pound  must  he  sell  it  to  gain  12^  per 
cent? 

4.  For  how  much  per  yard  must  cloth,  costing  $2.50  per 
yard,  be  sold,  to  gain  10  per  cent  on  its  cost? 


•  When  the  given  per  cent  can  be  reduced  to  a  convenient  vulgar  frac- 
tion, as  in  this  example,  it  is  ordinarily  best  to  use  the  vulgar  fraction 
instead  of  the  given  per  cent. 


PROFIT    AND    LOSS.  317 

5.  1  bought  cloth  at  $2  per  yard,  and  sold  it  at  $2.33  per 
yard.     What  per  cent  of  its  cost  did  I  gain  ? 

Solution.  —  Since  I  gave  $2  per  yard  for  the  cloth,  and  received 
$2.33,  I  received  $.33  per  yard  more  than  I  gave ;  and  as  $.33  =  ]^n> 
of  $2,  my  gain  must  be  ^^xnr  =  16^  per  cent  of  the  cost. 

6.  What  per  cent  shall  I  lose  by  selling  molasses,  which 
cost  me  $.30  per  gallon,  for  $.25  per  gallon  ? 

Solution. —  Since  I  bought  the  molasses  for  30  cents,  and  sold  it  for 
25  cents,  per  gallon,  I  lost  5  cents  per  gallon  on  it ;  and,  as  5  cents  =* 
;^  =  ig-  of  30  cents,  I  lost  ^,  or  16§  per  cent  of  the  cost. 

7.  How  much  per  cent  shall  I  gain  by  selling  flour,  which 
cost  me  $6.50  per  barrel,  for  $7,215  per  barrel  ? 

8.  Mr.  Winsor  bought  flour  at  $7.50,  and  sold  it  nt  $6.75 
per  barrel.     How  much  per  cent  did  he  lose  ? 

9.  What  per  cent  shall  I  gain  by  selling  linen  cloth,  which 
cost  45  cents  per  yard,  at  50  cents  per  yard  ? 

10.  What  per  cent  shall  I  lose  by  selling  linen  cloth,  cost- 
ing 50  cents  per  yard,  at  45  cents  per  yard  ? 

11.  A  merchant  bought  some  molasses  at  $.20  per  gallon, 
and  some  oil  at  $1.16  per  gallon.  He  sold  the  molasses  at 
$.25  per  gallon,  and  *the  oil  at  such  rate  that  he  gained  tlie 
same  per  cent  on  it  that  he  gained  on  the  molasses.  For 
how  much  per  gallon  did  he  sell  the  oil  ? 

12.  I  gained  $17.28  by  selling  a  lot  of  sugar  for  16  per 
cent  more  than  it  cost.  How  many  dollars  did  it  cost,  and  for 
how  many  did  I  sell  it  ? 

Suggestion. —  Since  I  gained  16  per  cent,  or  ^  of  the  cost.  I  must 
have  sold  it  for  f  f  of  the  cost.  Hence,  the  cost  =  W  and  the  price 
for  which  I  sold  it  =  -^-  of  the  gain. 

13.  Mr.  Kent  bought  a  lot  of  apples,  and  sold  them  for  20 
per  cent  more  than  they  cost,  by  which  he  gained  $24.80. 
How  much  did  they  cost  him,  and  for  how  much  did  he  sell 
them  ? 

14.  Mr.  Kilburn  sold  43  barrels  of  apples  for  $6.45  less 
than  they  cost  him,  thereby  losing  10  per  cent  of  their  cost. 

27*  , 


318  PROFIT    AND    LOSS. 

What  did  they  cost  him,  and  what  did  he  get  per  barrel  fof 
them  ? 

10.  Mr.  Thurbur  sold  146  yards  of  cloth  for  $71.54  more 
than  it  cost  him,  thereby  gaining  14  per  cent.  How  much 
did  he  receive  per  yard  for  it  ? 

1 6.  Logee  &  Drown  sold  a  large  lot  of  goods  for  $6700.43^, 
thereby  gaining  18  per  cent  on  their  cost.  How  much  did 
the  goods  cost  them  ? 

Suggestion.  —  Since  they  gained  18  per  cent  of  the  cost,  $6700.43^ 
must  be  118  per  cent,  or  ff  of  the  cost. 

17.  What  is  the  cost  of  a  lot  of  goods  on  which  15  per 
cent  will  be  gained  by  selling  them  for  $288.65  ? 

18.  What  is  the  cost  of  a  lot  of  goods  on  which  8  per  cent 
will  be  gained  by  selling  them  for  $622,215  ? 

19.  What  is  the  cost  of  a  lot  of  goods  on  which  7  per  cent 
will  be  lost  by  selling  them  for  $442.68  ? 

Suggestion.  —  Since  7  per  cent  will  be  lost  by  the  sale,  $442.68  must 
equal  93  per  cent  of  the  cost. 

20.  What  is  the  cost  of  a  lot  of  goods  on  which  30  per 
( 3nt  will  be  lost  by  selling  them  for  $874,846  ? 

21.  What  is  the  cost  of  a  lot  of  goods  on  which  9  per  cent 
^  «rill  be  lost  by  selling  them  for  $9009  ? 

22.  A  man  bought  corn  at  50  cents  per  bushel,  for  which 
iie  asked  25  per  cent  more  than  it  cost  him  ;  but  it  falling  in 
price,  he  was  obliged  to  sell  it  for  25  per  cent  less  than  hia 
asking  price.  Did  he  gain  or  lose,  and  how  much  per  cent  ? 
How  much  on  each  bushel  ? 

Solution.  —  Since  his  asking  price  was  12.5  per  cent,  or  ^  of  the  cost, 
and  his  selling  price  was  75  per  cent,  or  f  of  liis  asking  price,  his  selling 
price  must  have  been  f  of  f-,  or  \^  of  tlie  cost.  Therefore  he  lost  yV* 
or  65-  per  cent  of  the  cost ;  and  since  each  bushel  cost  $.50,  his  loss  per 
bushel  must  have  been  r\j  of  $.50,  which  is  $.03^. 

23.  A  merchant  asked  for  a  lot  of  goods  1 2^  per  cent  more 
than  they  cost  him,  but  was  obliged  to  deduct  12^  per  cent 
from  his  asking  price.     What  part  of  the  cost  did  he  lose  ? 

24.  I  asked  for  a  lot  of  cloth  1 0  per  cent  more  than  it  cost 


PROFIT    AND    LOSS.  31  J* 

rrs,  but  was  obliged  to  deduct  10  per  cent  from  my  asking 
price.     What  per  cent  of  its  cost  did  I  lose  ? 

25.  A  watch  dealer  bought  a  watch  for  $75,  and  asked  for 
it  33^  per  cent  more  than  it  <iost.  He  was  obliged  to  sell  it 
for  1 0  per  cent  less  than  his  asking  price.  What  per  cent  did 
he  gain  on  the  investment  ?     How  many  dollars  ? 

26.  A  merchant  asked  for  a  quantity  of  goods  25  per  cent 
more  than  they  cost  him,  but  was  obliged  to  sell  them  for  12^ 
per  cent  less  than  his  asking  price.  He  gained  $98.70  by  the 
transaction.  How  much  did  the  goods  cost  ?  For  how  much 
did  he  sell  them  ?     What  was  his  asking  price  ? 

Suggestion.  —  His  asking  price  was  f  of  the  cost,  and  his  selling  price 
was  ^  of  his  asking  price,  or  |-  of  :f  ==  %^  of  the  cost.  Hence,  the 
gain,  $98.70,  =  -js"  of  the  cost. 

27.  A  merchant  asked  for  a  lot  of  cloth  20  per  cent  more 
than  it  cost  him  ;  but  being  in  want  of  money,  he  sold  the 
cloth  for  25  per  cent  less  than  his  asking  price.  Allowing 
that  he  lost  $.275  per  yard,  for  how  much  per  yard  did  he 
sell  it? 

28.  I  sold  a  lot  of  goods  for  $4268,  which  was  8^  per  cent 
less  than  my  asking  price.  My  asking  price  was  44  per  cent 
more  than  the  cost  of  the  goods.     How  much  did  they  cost  ? 

29.  By  selling  flour  at  $6.65  per  barrel,  I  shall  lose  5  per 
cent  of  its  cost.  For  how  much  per  barrel  must  I  sell  it  to 
gain  5  per  cent  ? 

Suggestion.  —  $6.65  =  95  per  cent  of  the  cost.  Hence,  105  per  cent 
of  the  cost  must  equal  -^5^-  of  $6.65. 

30.  By  selling  land  at  $88.14  per  acre  I  shall  gain  13  per 
cent.     For  how  much  must  I  sell  it  to  gain  20  per  cent  ? 

31.  By  selling  a  lot  of  goods  for  $113.75  I  lost  9  per  cent. 
For  how  much  ought  I  to  have  sold  it  to  gain  9  per  cent  ? 

32.  I  bought  400  bushels  of  grain,  and  sold  half  of  it  at 
75  cents  per  bushel,  which  was  20  per  cent  more  than  it  cost. 
I  sold  the  remainder  at  an  advance  of  25  per  cent  on  the  cost. 
For  how  much  per  bushel  did  I  sell  the  last  lot  ?  How  many 
dollars  did  I  gain  ? 


320  PROFIT   AND    LOSS. 

33.  What  must  be  the  asking  price  of  cloth  costing  $2.55 
per  yard,  that  I  may  fall  10  per  cent  on  it,  and  still  gain  14 
per  cent  on  the  cost  ? 

Suggestion.  —  As  the  selling  price  is  to  be  90  per  cent,  or  -jVd  of  the 
asking  price,  and  114  per  cent  of  the  cost,  the  asking  price  must  be  -^^j^ 
of  the  selling  price,  and  "Vrj*-  of  \^^  =  -Vir  of  the  cost. 

34.  What  must  be  the  asking  price  of  cloth  costing  $3.29 
per  yard,  that  I  may  deduct  12^  per  cent  from  it,  and  still 
gain  12|^  per  cent  on  the  cost  ? 

35.  What  must  be  the  asking  price  of  boots  costing  $2.75 
per  pair,  that  I  may  fall  16f  per  cent  on  it,  and  still  gain  20 
per  cent  on  their  cost  ? 

36.  If,  by  selling  cloth  at  $2.34  per  yard,  I  gain  4  per  cent 
of  its  cost,  what  per  cent  shall  I  gain  by  selling  it  at  $*2.79 
per  yard  ? 

Suggestion  and  Answer.  —  Since  $2.34  =  4  per  cent  more  than  the 
cost,  it  must  equal  104  per  cent  of  the  cost.  Hence,  $2.79  =  f  ^|  of 
104  per  cent,  =  124  per  cent  of  the  cost,  and  the  gain  equals  24  per 
cent. 

37.  If,  by  selling  cloth  at  $4.37  per  yard,  I  gain  15  per 
cent  of  its  cost,  what  per  cent  shall  I  gain  by  selling  it  at 
$4.94  per  yard  ? 

38.  If,  by  selling  flour  at  $6.75  per  barrel,  I  gain  8  per 
cent  of  its  cost,  what  per  cent  shall  I  lose  by  selling  it  at  $6 
per  barrel  ? 

39.  If,  by  selHng  oil  at  $1,254  per  gallon,  I  lose  5  per  cent 
of  its  cost,  what  per  cent  shall  I  gain  by  selling  it  at  $1,584 
per  gallon  ? 

40.  I  bought  96  acres  of  land  at  $84  per  acre,  and  sold  ^ 
of  it  for  the  cost  of  the  whole.  What  per  cent  did  I  gain  on 
the  part  sold  ? 

Suggestion.  —  The  part  sold  cost  |-,  and  was  sold  for  |  of  the  cost  of 
the  entire  lot.    Hence,  it  was  sold  for  §  of  its  cost. 

41.  I  bought  28  tons  of  iron  at  $48  per  ton,  and  sold  f  of 
it  for  the  cost  of  the  whole.  What  per  cent  did  I  gain  on  the 
part  sold  ? 


I 

PROFIT    AND    LOSS.  321 

42.  I  bought  83  barrels  of  beef  at  $12.50  per  barrel,  and 
was  obliged  to  sell  it  for  what  ^  of  it  cost.  What  per  cent 
did  I  lose  ? 

43.  A.  Sh  M.  J.  Miles  bought  of  Cragin  &  Cleveland,  for 
cash,  goods  to  the  amount  of  $423.75,  and  the  same  day  sold 
them  at  an  advance  of  16  per  cent,  receiving  in  payment  a 
note  on  3  months.  This  note  they  got  discounted  at  a  bank  at 
the  rate  of  6  per  cent  per  year.  How  much  did  they  gain  on 
the  goods  ? 

44.  Armington,  Horswell,  &  Kilburn  bought  of  Godding, 
Briggs,  &  Co.  goods  to  the  amount  of  $1000,  })ayabl6  in  6 
months,  without  grace.  One  month  afterwards  they  sold  the 
goods  for  cash,  at  an  advance  of  10  per  cent,  and  immediately 
put  the  money  at  interest  at  6  per  cent.  When  the  6  months 
had  expired,  they  collected  the  amount  of  the  money  they  had 
lent,  and  paid  the  bill  due  Godding,  Briggs,  &  Co.  Did  they 
gain  or  lose,  and  how  many  dollars  ? 

45.  I  bought  a  lot  of  coffee  at  12  cents  per  pound.  Allow- 
ing that  5  per  cent  of  the  coffee  will  waste  in  weighing  it  out, 
and  that  10  per  cent  of  the  sales  will  be  bad  debts,  for  how 
much  per  pound  must  I  sell  it  to  make  a  clear  gain  of  14  per 
cent  on  the  cost  ? 

Answer.     16  cents  per  pound. 

46.  What  must  be  the  asking  price  of  raisins  costing  $7.29 
per  cask,  that  I  may  fall  10  per  cent  from  it  and  still  gain  10 
per  cent  on  the  cost,  allowing  that  10  per  cent  of  the  sales 
will  be  bad  debts  ? 

47.  I  bought  8  casks  of  oil,  each  containing  133  gallons, 
at  $1.20  per  gallon,  and  paid  $5.32  for  having  it  brought  to 
my  store.  Allowing  that  there  will  be  a  waste  of  5  per  cent 
in  measuring,  that  3  per  cent  of  my  sales  will  be  bad  debts, 
and  that  it  will  cost  1  per  cent  of  the  remainder  to  collect  itj 
for  how  much  per  gallon  must  I  sell  it  to  make  a  net  gain  oi 
33  per  cent  on  its  cost  at  my  store,  nothing  being  allowed  foi 
interest  ? 


322  PARTXKRSHIP. 

^00.    Partner- ship. 

Two  or  more  persons,  uniting  for  the  purpose  of  carrying 
on  business  together,  form  what  is  called  a  partnership, 
FIRM,  or  COMPANY.  The  capital  invested  by  them  is  called 
their  stock  in  trade. 

It  is  evident  that  the  profit  or  loss  made  by  the  company  should  be 
shared  among  its  members  in  proportion  to  what  the  use,  or  interest,  of 
each  man's  stock  for  the  time  it  was  invested  is  worth. 

When  the  stocks  of  the  several  partners  are  invested  for  the  same 
lengrth  of  time,  their  use,  or  interest,  will  be  proportioned  to  the  stocks 
themselves,  and  hence  each  partner's  gain  or  loss  will  be  the  same  part 
of  liis  stock  that  the  entire  gain  or  loss  is  of  the  entire  stock ;  or  it  will 
be  the  same  part  of  the  entire  gain  or  loss  that  his  stock  is  of  the  entire 
stock. 

1.  A,  B,  and  C  trade  in  company.  A  puts  in  $250,  B 
puts  in  $750,  and  C  puts  in  $500.  At  the  end  of  6  months 
they  find  that  they  have  gained  $472.50.  What  is  each  man's 
share  of  the  gain  ? 

First  Solution.— Since  A's  stock  =  $250,  B's  =  $750,  and  C's  =^ 
$500,  the  entire  stock  =  $250  -}-  $750  +  $.500  =  $1500;  and  as  the 
gain  =  $472.50,  it  must  equal  j^uwiju,  or  ^jj  of  the  stock.  There- 
fore, each  man's  gain  will  be  ^^jj  of  his  stock,  which  gives  for  A's 
gain  $78.75,  for  B's  gain  $236.25,  and  for  C's  gain  $157.50. 

Second  Solution.  —  Since  A's  stock  =  $250,  B's  =  $750,  and  C's  = 
$500,  the  entire  stock  must  equal  $1500,  of  which  A's  stock  =  -it^ 
=  ^,  B's  =  /^jj"^  =  i,  C's  =  iWiy  =  h  Therefore,  A  should  have 
^,  B  should  have  ^,  and  C  should  have  ^  of  the  gain.  ^  of  $472.50 
=  $78.75  =  A's  share ;  ^  of  $472.50  =  $236.25  =  B's  share ;  i  of 
$472.50  =  $157.50  =  C's  share. 

2.  X,  Y,  and  Z  traded  in  company  for  1  year.  X  put  in 
$1000,  Y  put  in  $1500,  and  Z  put  in  $2000.  At  the  end  of 
the  year  they  found  that  they  had  gained  $1800.  What  was 
each  man's  share  of  the  gain  ? 

3.  A  man,  failing  in  business,  finds  that  he  owes  A  $424, 
B  $638,  C  $197,  D  $338,  and  E  $574,  and  that  his  whole 
available  property  amounts  only  to  $1173.  How  much  ought 
he  to  pay  to  each  creditor  ? 


PARTNERSHIP. 


323 


Suggestion. —  Since  he  owes  $2171,  and  has  but  $1173,  he  can  pay 
but  srff  of  his  debts.  Therefore,  he  ought  to  pay  A  irf f  of  $424, 
B  ^f  of  $638,  &c. 

4.  The  stock  of  a  bankrupt  is  valued  at  $1200,  and  he 
owes  $4200.  How  many  dollars  ought  he  to  pay  the  person 
to  whom  he  owes  $546  ?   to  whom  he  owes  $338.73  ? 

5.  A,  B,  C,  and  D  agree  to  cut  500  cords  of  wood  for 
$300.  When  the  job  is  finished,  they  find  that  A  has  cut  125 
cords,  B  100  cords,  C  150  cords,  and  D  the  rest.  How  many 
dollars  ought  each  to  receive  ? 

6.  A  and  B  traded  in  company.  A  put  in  $200,  and  B 
put  in  $300.  A's  share  of  the  gain  was  $84.56.  What  was 
B's  share  ? 

7.  A  and  B  traded  in  company,  and  gained  $348,  of  which 
B's  share  was  $261.  If  A's  stock  was  $175,  what  was  B's 
stock,  and  A's  share  of  the  gain  ? 

8.  Samuel  Greene  and  Joseph  Irons  traded  in  company, 
Greene  paid  in  3  times  as  much  of  the  stock  as  Irons,  and 
they  gained  $1176.     What  was  each  one's  share  of  the  gain  ? 

Suggestion.  —  Since  Greene  paid  in  3  times  as  much  as  Irons,  both 
together  must  have  paid  in  4  times  as  much  as  Irons.  Therefore,  Irons 
paid  in  5^,  and  Greene  f  of  the  stock. 

9.  William  Balch  and  Joseph  Adams  bought  a  ship  to- 
gether, Balch  paying  in  twice  as  much  money  as  Adams.  At 
the  end  of  one  year  they  sold  her,  and  found  that  they  had 
realized  a  profit  of  $15,000  from  her.  What  was  each  part- 
ner's share  ? 

10.  Anderson  and  Pai'ker,  after  trading  in  company  for  2 
years,  found  that  their  profits  had  been  $2400.  Allowing 
that  Anderson's  stock  was  f  of  Parker's,  how  many  dollars 
of  the  profit  ought  each  to  have  ? 

11.  A,  B,  and  C  traded  in  company.  A  put  in  ^  of  the 
stock,  B  put  in  ^  of  it,  and  C  put  in  the  rest.  On  dividing 
the  gain,  they  found  that  C's  share  of  it  was  $321.  What 
was  the  train  of  each  of  the  other  partners  ? 

12.  William  Hall,  Edward  Johnson,  and  Henry  Whiting 


324  .  PAltTNEKSHIP    ON    TIME. 

traded  in  company,  and  gained  $6534,  of  which  Johnson's 
share  was  $1089.  If  Johnson's  and  Whiting's  stock  was 
together  equal  to  twice  Hall's,  what  was  Hall's  share  of  the 
gain  ?     What  was  Johnson's  share  ? 

13.   A  small  estate  belonged  to  a  large  number  of  heirs : 

2  members  of  the  family  of  A  each  owned  y-^^  of  the  estate  ; 
4  of  the  family  of  B  each  owned  ^^^  of  it ;  4  of  the  familj 
of  C  each  owned  ^  of  it ;  2  of  the  family  of  D  each  owned 
jj\j  of  it ;  4  of  the  family  of  E  each  owned  -^^  of  it ;  3  of  the 
family  of  F  each  owned  -^f-g-  of  it ;  4  of  the  family  of  G  each 
owned  -^^-^  of  it ;  6  of  the  family  of  H  each  owned  yf-^  of  it ; 

3  of  the  family  of  I  each  owned  -^^  of  it.  Mr.  Byram,  as 
agent  for  the  above-named  individuals,  sold  their  interest  in 
the  estate  for  $350.     How  many  dollars  ought  he  to  give  to 

each  ? 

Ansioer. 

$  3.123  to  each  member  of  A's  family. 
$  1.562  to  each  member  of  B's  family. 
$31,234  to  each  member  of  C's  family. 
$15,617  to  each  member  of  D's  family. 
$12,493  to  each  member  of  E's  family. 
$  1.785  to  each  member  of  F's  family. 
$  2.499  to  each  member  of  G's  family. 
$  8.924  to  each  member  of  H's  family. 
$20,823  to  each  member  of  I's  family. 

Note. —  The  above  example  is  a  statement  of  transactions  which 
actually  occurred.  It  was  brought  to  the  author  for  solution,  by  th*» 
agent  of  the  parties. 


910.    Partnership  on  Time. 

In  dividing  the  gain  or  loss  among  the  partners,  when  their 
shares  of  the  stock  are  invested  for  unequal  times,  it  becomes 
necessary  to  consider  both  the  stock  and  the  time,  or  to  con- 
sider the  interest  of  each  man's  stock  for  the  time  it  was  in 
trade.  The  following  examples  and  solutions  will  illustrate 
this  :  — 

1    A,  B,  and  C  traded  in  company.     A  put  in  $750  for  10 


PARTNERSHIP    ON    TIME.  325 

months,  B  put  in  $375  for  12  months,  and  C  put  in  $1125 

for  16  months.     They  gained  $860.     What  was  each  man's 

rV&ve  of  the  gain  ? 

First  Solution. 

$37.50  =  interest  of  A's  stock  for  10  mo. 
22.50  =  interest  of  B's  stock  for  12  mo. 
90.00  =  interest  of  C's  stock  for  16  mo. 


$150.00  =  interest  of  whole. 
Therefore,  A  should  have  yVtsi/iTj  or  5-,   of  the  gain,  =  $215. 
B  should  have  -Y^xyuxf-,  or  ^tr>  of  the  gain,  =  $129. 
.     C  should  have  tWA,  or  f ,    of  the  gain,  =  $516. 

Second  Solution. 
The  use  of  $  750  for  10  mo.  is  worth  the  use  of  $  7500  for  1  mo. 
The  use  of  $  375  for  12  mo.  is  worth  the  use  of  $  4500  for  I  mo. 
The  use  of  $1125  for  16  mo.  is  worth  the  use  of  $18000  for  1  mo. 


Use  of  whole  stock  is  worth  the  use  of  $30000  for  1  mo. 
Therefore,  A  should  have  -^wu^^,  or  ^,    of  the  gain,  =  $215. 
B  should  have  -suijGu,  or  -^jj,  of  the  gain,  =  $129. 
C  should  have  it^^§,  or  f,    of  the  gain,  =  $516. 
When  the  stocks  of  the  several  partners  are  convenient  multiples 
or  fractional  parts  of  each  other,  a  very  neat  solution  can  be  given. 
Thus,  in  the  above  example,  by  noticing  that  B's  stock  equals  ^  of  A's, 
and  that  C's  stock  equals  |  of  A's,  we  may  have  the  following :  — 

TTiird  Solution.  —  The  use  of  A's  stock  10  mo.  =  use  of  10  times 
A's  stock  for  1  mo. 

The  use  of  B's,  or  ^  of  A's  stock,  12  mo.  =  use  of  ^,  or  6  times 
A's  stock  for  1  mo. 

The  use  of  C's,  or  f  of  A's  stock,  16  mo.  =  use  of  ^-,  or  24  times 
A's  stock  for  1  mo. 

Use  of  whole  =  use  of  10  +  6  +  24,  or  40  times  A's  stock  for 
1  mo. 

Therefore  A  should  have  |^,  or  ^  ;  B  /^y,  or  ^ ;  and  C  f^,  or  f , 
of  the  gain,  which  will  give  the  same  answer  as  before. 

2.  Charles  French,  Francis  Baker,  and  Otis  Atherton 
traded  in  company,  under  the  name  of  Charles  French  &  Co. 
French  put  in  $1000  for  20  mo.,  Baker  put  in  $800  for  16 
mo.,  and  Atherton  put  in  $500  for  20  mo.  They  gained 
$1500.  How  many  dollars  of  the  gain  ought  each  to  receive  ** 
28 


826  PAETNEESHIP    ON   TIME. 

3.  George  Jackson,  William  Leach,  and  Albert  Buflington 
traded  in  company.  Jackson  put  in  $144  for  6  mo.,  Leach 
put  in  $72  for  7  mo.,  and  Buffington  put  in  $216  for  6  mo. 
20  da.  They  gained  $114.  What  was  each  man's  share  of 
the  gain  ? 

4.  A,  B,  C,  and  D  hired  a  pasture  together,  in  which  A 
pastured  4  cows  13  weeks,  B  pastured  5  cows  16  weeks,  C 
pastured  8  cows  10^  weeks,  and  D  pastured  4  cows  16  weeks. 
The  rent  of  the  pasture  was  $102.  How  many  dollars  ought 
each  man  to  pay  ? 

5.  Samuel  Austin,  Jacob  Brown,  and  Moses  Sumner 
formed  a  partnership  for  2  years,  under  the  name  of  Samuel 
Austin  &  Co.  Austin  at  first  paid  in  to  the  stock  $1000,  but 
after  8  mo.  had  elapsed  he  paid  in  $500  more.  Brown  at 
first  paid  in  $1250,  and  16  mo.  afterwards  he  paid  in  $250 
more.  Sumner  at  first  paid  in  $1500,  but  at  the  end  of  16 
mo.  he  took  out  $500.  They  gained  $3600.  What  was  each 
man's  share  of  the  gain  ? 

Solution. 

Interest  of  $1000  for  8  mo.    =$40)         -_^       .  ^    .      ^.  ,     ^    , 
T  z.  *,  r^^  r     ,  ^  *,  .^^  }•  =  ^160  =  int.  Austin's  stock. 

Interest  of  $1500  for  16  mo.  =  $120  ) 

Interest  of  $1250  for  16  mo.  =  $100  )         .,^^       .  ^  _,         ,     ^    . 

-^,^„„  ^  „^\  =  $160  =  mt.  Brown's  stock. 

Interest  of  $1500  for  8  mo.    =  $  60  ) 

Interest  of  $1500  for  16  mo.  =  $120  )         «.,^«       •      c  ,  i 

T  *       ^    ruMnnnf     o  *  ^/^  f  =  $160  =  int.  Sumncr's  stock. 

Interest  of  $1000  for  8  rao.    =  $  40  j 

By  this,  it  appears  that  the  interests  of  their  respective  stocks,  for  the 
time  they  were  in  trade,  were  alike.  Hence,  the  gain  should  be  divided 
equally,  and  each  partner  should  have  ^  of  $3600,  which  is  $1200. 

Note.  —  Other  solutions  similar  in  character  to  those  given  to  the 
first  example  might  have  been  added ,  hut  as  the  pupil  can  readily  dis- 
cover them,  they  have  been  omitted. 

6.  Joseph  Southwick,  Francis  Lowe,  and  Henry  Taft 
formed  a  partnership  for  3  years,  under  the  name  of  South- 
wick, Lowe,  &  Taft.  When  they  commenced  business,  each 
partner  put  in  $3000 ;  but  at  the  end  of  the  first  year  South- 
wick put  in  $3000  more,  and  Lowe  withdrew  $1500.  At  the 
end  of  the  second  year,  Southwick  withdrew  $2000,  and 


POWERS    AND    ROOTS.  327 

Lowe  put  in  $4000,  and  Taft  put  in  $2000.  When  the  part- 
nership expired,  they  found  that  they  had  gained  $9000. 
What  was  each  partner's  share  of  the  gain  ? 

7.  S.  Gamwell,  C.  Grover,  R.  Wheelock,  and  W.  Godding 
formed  a  partnership,  under  the  title  of  Gamwell,  Grover,  & 
Co.  Gamwell  at  first  put  in  $8000,  but  at  the  end  of  6  mo. 
he  withdrew  $2000,  and  at  the  end  of  12  mo.  he  withdrew 
$1000  more.  Grover  at  first  put  in  $6000,  but  at  the  end 
of  10  mo.  he  put  in  $3000  more.  Wheelock  put  in  $7000. 
Godding  at  first  put  in  $10,000  ;  at  the  end  of  6  mo.  he  with- 
drew $2000,  and  at  the  end  of  14  mo.  he  put  in  $4000.  At 
the  end  of  2  years  they  found  that  they  had  gained  $12,000. 
What  was  each  man's  share  of  the  gain  ? 


SECTION   XVI. 
POWERS  AND   ROOTS. 

311.    Definitions. 

(a.)  The  product  of  a  number  taken  any  number  of  times 
as  a  factor  is  called  a  power  of  the  number.  —  See  10^, 
{d.)   {e.)   (/.)  and  Note. 

(5.)  A  ROOT  of  any  number  is  such  a  number  as,  taken 
some  number  of  times  as  a  factor,  will  produce  the  given 
number. 

(c.)  If  the  root  must  be  taken  twice  as  a  factor  to  produce 
the  number,  it  is  the  square  root,  or  the  second  root  ; 
if  three  times,  it  is  the  cube  root,  or  the  third  root  ;  if 
four  times,  it  is  the  fourth  root  ;  &c. 

Thus,  2  is  the  square  root  of  4,  the  third  root  of  8,  the  fourth  root  of 
16,  &c.,  because  22  =  4,  23  =  8,  2*  =  16  ;  &^. 

(rf.)    The  character  ^,  called  the  radical  sign,  is  used 


828       RELATION  OF  A  SQUARE  TO  ITS  ROOT. 

to  indicate  that  the  root  of  the  number  over  which  it  is  placed 
is  to  be  extracted. 

(e.)  The  degree  of  the  root  is  indicated  by  a  small  fig- 
ure, called  an  index,  which  is  placed  a  Kttle  above  and  at  the 
left  of  the  sign.  When  no  index  is  written,  the  square  root 
is  required. 

Thus,  ^  4,  or  ^  4,  means  the  square  root  of  4. 

jyiiS  means  the  fifth  root  of  243. 

^  7^  means  the  fourth  root  of  the  3d  power  of  7. 

(/')  ^®  ^^y  ^^^0  indicate  that  a  root  is  to  be  extracted, 
by  using  a  fractional  exponent. 

Thus,  92  =  ^9";    (125)' =  ^"125;   27?  =  ^272;    &c. 

(g.)  The  process  of  finding  the  powers  of  numbers  is 
called  Involution,  and  the  process  of  finding  their  roots  is 
called  Evolution,  or  the  Extracting  of  RbOTS. 


313.  Melation  which  the  Denominations  of  a  Square  bear 
to  those  of  its  Root, 

(a.)    TABLE   OP   SQUARES. 
12=1  42=16  72  =  49 

22  =  4  52  =  25  82  =  64 

32  =  9  62  =  36  92=81 

1 02  =  1 00  1 0,0002  _  1 ,000,000,000 

1002  =  10,000  100,0002  =  10,000,000,000 

1,0002  =  1,000,000  1,000,0002  =  1,000,000,000,000 

(6.)  The  above  table  shows  that,  —  Eirst.  There  are  below  100  but  9 
entire  numbers  which  are  perfect  squares. 

Second.  The  entire  part  of  the  square  root  of  any  number  below  100 
will  be  less  than  10,  and  therefore  contain  but  1  figure;  of  any  num- 
ber between  100  and  10,000  will  lie  between  10  and  100,  and  therefore 
contain  2  figures;  between  10,000  and  1,000,000  will  lie  between  100 
and  1000,  and  therefore  will  contain  3  figures  ;  &c. 

1.  How  many  figures  are  there  in  the  entire  part  of  the 
square  root  of  865698  ? 

Answer.  —  Since  865698  lies  between  10,000  and  1,000,000,  iU  root 


DIVISION    INTO    PERIODS.  329 

must  lie  between  the  roots  of  those  number?,  i.  c,  between  100  and  1000, 
and  must  therefore  contain  3  figures  in  its  entire  part. 

How  many  figures  are  there  in  the   entire   part  of  the 
root  of — 


2.  69748769  ? 

3.  486497950068  ? 


4.  12496743297  ? 

5.  5847695329  ? 


313.    Division  into  Periods, 

(a.)  As  the  square  of  10  is  100,  of  100  is  10,000,  &c.,  it  follows  that 
the  square  of  any  number  of  tens  will  be  some  number  of  hundreds ; 
of  any  number  of  hundreds  will  be  some  number  of  ten  thousands,  &c. ; 
or,  in  other  words,  that  the  square  of  tens  will  give  units  of  no  denomi- 
nation below  hundreds  ;  the  square  of  hundreds  will  give  units  of  no 
denomination  below  ten-thousands  ;  &c. 

{h.)  Hefice,  the  two  right  hand  figures  of  any  number  will  contain 
no  part  of  the  square  of  the  denominations  of  the  root  above  units ;  the 
four  right  hand  figures  will  contain  no  part  of  the  square  of  those  above 
tens,  &c. 

(c.)  Therefore,  if  we  should  begin  at  the  right  of  any  number,  and 
separate  it  into  periods  of  two  figures  each,  the  number  of  periods  would 
be  the  same  as  the  number  of  figures  in  its  square  root.  The  square  of 
the  highest  denomination  of  the  root  would  be  found  in  the  left  hand 
period  ;  the  square  of  the  two  highest  denominations  would  be  found  in 
the  two  left  hand  periods  ;  &c. 

1.    Separate  8478695  into  periods,  and  explain  their  uses. 

Anncer.  8478695.  The  left  hand  period,  8,  contains  all  of  the  square 
of  the  thousands  of  the  root ;  the  two  left  hand  periods,  847,  the  square 
of  the  thousands  and  hundreds  ;  &c. 

Separate  each  of  the  following  numbers  into  periods,  and 
explain  their  uses  :  — 

2.  5794865.  I  4.     375486792. 

3.  89475948.  5.     32500675. 


31^1:.    Method  of  forming  a  Square. 

(a  )    To  find  a  law  of  universal  application  in  squaring  or  extracting 
the  square  roots  of  numbers,  we  will  use  the  letter  a  to  represent  any 
anmber  whatever  and  1)  to  represent  any  other  number. 
28* 


830  METHOD    OF    FORMING    A    SQUARE. 

(ft.)  Then  will  a  -f-b  represent  the  sum,  and  (a  -\-  b)2  or  (a  -|-  b) 
X  (a-  -|"  b)  the  square  of  the  sum  of  any  two  numbers  whatever. 

(c.)  Performing  the  multiplication,  we  have  a  times  a  =  a2;  a  times 
b  =  a  X  b,  or,  as  it  mav  be  written,  ab ;  b  times  a  =  a  times  b  =  a 
X  b,  or  a  b  ;  b  times  b  =  b2. 

(d.)  Writing  the  work,  a.s  below,  and  adding  the  partial  products,  wo 
have,  — 

a  +b 
a  +b 

a  X  (a  4-  b)  =  a2  -f-  a  times   b  =  a2  -f-  ab 

bX(a-|-b)=      -f- a  times   b  +  b2  =  ab  +  b* 

(a  +  b)  X  (a  +  b)  =  a2  -f-  2  times  ab  +  b2  =  a2  -f  2  ab  +  ba 

(e.)  Hence,  (a  +  b)2  =  a2  -j-  2  ab  +  b2,  or,  since  a2  equals  the 
square  of  the  first  number,  and  2  ab  equals  twice  the  product  of  the 
first  number  by  the  second,  and  b2  equals  the  square  of  the  second ; 

The  square  of  the  sum  of  any  two  numbers  equals  the  square  of  the  first, 
plus  tioice  the  product  of  the  first  by  the  second,  plus  the  square  of  the 
second. 

Illustrations. 

(7  4-  5)2  =  72  4-  2  X  7  X  5  +  52  =  49  +  70  +  25  =  144  =  122 
(8  -i-  4)2  =  82  4-  2  X  8  X  4  -i-  42  =  64  -f  64  -i-  16  =  144  =  122 
(20  4-  3)2  =  202  +  2  X  3  X  20  -f  32  =  400  +  120  4-  9  =  529  =  232 

(/)  But  a2  4-  2  ab  4-  b2  can  be  put  into  another  form  ;  for  2  ab  4- 
b2  =  2  a  times  b  4-  b  times  b,  =  (2  a  4-  b)  times  b,  or  (2  a  4-  b)  X  b, 
or,  by  omitting  the  sign  Xj  as  may  be  done  without  ambiguity,  (2  a  4- 
b)  b. 

Hence,  (a  4-  b)2  =  a2  4-  2  ab  4-  b2  =  a2  4-  (2  a  4-  b)  b. 

(g.)  But  a2  means  the  square  of  the  first  number ;  2  a  4-  b  means  the 
sum  of  twice  the  first  number,  plus  the  second ;  and  (2  a  4-  b)  b  the 
sum  of  twice  the  first,  plus  the  second,  multiplied  by  the  second. 

(A.)  JIe7ice,  the  square  of  the  su7n  of  two  numbers  is  also  equal  to  the 
square  of  the  first  number,  plus  the  product  obtained  by  multiplying  the  sum 
of  twice  the  first  number  plus  the  second,  by  the  second. 

Illustrations. 
(7  4.  5)2  =  72  4-  (14  4-  5)  5  =  49  4-  95  =  144  =  122. 
(8  -i-  4)2  =  82  4-  (16  +  4)  4  =  64  4-  80  =  144  =  122. 
(40  4-  8)2  =  402  4-  (80  +  8)  8  =  1600  4-  704  =  2.304  =  482. 

(t.)  Now,  as  any  number  above  ten  is  composed  of  tens  and  units, 
its  square  will  be  composed  of  the  square  of  the  tens,  plus  the  prodact 
of  twice  the  tens  plus  the  units  multiplied  by  the  units. 

(j.)   If  there  are  more  than  ton  tens  in  the  number,  the  part  which  ii 


METHOD  OF  EXTRACTING  THE  SQUARE  ROOT.   331 

composed  of  tens  may  be  considered  as  made  up  of  hundreds  and  tens, 
and  its  square  will  equal  the  square  of  the  hundreds,  plus  the  product 
of  twice  the  hundreds,  plus  the  tens,  multiplied  by  the  tens. 

(k.)  Proceeding  in  this  way,  we  shall  at  last  reach  the  part  which  is 
expressed  by  one  or  two  figures,  and  composed  of  only  the  two  highest 
denominations  of  the  given  number.  The  square  of  this  part  will  be 
the  square  of  the  highest  denomination,  plus  the  product  of  twice  the 
highest  denomination,  plus  the  next  lower,  multiplied  by  the  next  lower 
Thus,  — 

(4837)2  =  (4830  +  7)2  =  48302  -f  (2  X  4830  +  7  )  X  7 
(4830)2  =  (4800  4-  30)2  _  43002  -f  (2  X  4800  +  30)  X  30 
(4800)2  =  (4000  4-  800)2  =  40002  +  (2  X  4000  -f  800)  X  800 


2\3,    3fethod  of  exti'cicting  the  Sqiicrs  Root, 


What  is  the  value  of  V  925444  ? 

Solution. —  (a.)  Since  this  number  lies  between  10,000  and  1,000,000, 
its  root  must  lie  between  100  and  1000,  and  must  therefore  be  composed 
of  hundreds,  tens,  and  units.  Dividing  it  into  periods  of  two  figures 
each,  it  will  take  the  form  925444. 

(6.)  If,  now,  we  let  a  represent  the  hundreds  of  the  root,  and  b  the 
tens,  the  whole  of  a2  will  be  found  in  the  left  hand  period,  i.  e.,  in  the 
ten-thousands,  and  the  whole  of  (a  -f-  ^)^  in  the  two  left  hand  periods, 
i.  e.,  in  9254  hundreds. 

(c.)  The  greatest  square  in  92  is  81,  the  root  of  which  is  9.  There- 
fore, 9  =  a  =  the  hundreds  figure  of  the  root.  Subtracting  a2,  =  81 
ten-thousands,  from  92  ten-thousands,  leaves  11  ten- thousands,  to  which 
adding  the  54  hundreds  gives  1154,  which  must  contain  (2  a  -|-  b)  b. 

(d.)  Now,  as  we  know  a,  we  can  find  2  a,  and  make  use  of  it  as  a 
trial  divisor  to  find  b.  But  a  being  hundreds  and  b  tens,  2  ab  must  be 
thousands,  and  no  part  of  it  will  be  found  to  the  right  of  the  thousands. 

(e.)  Hence,  in  dividing,  we  may  disregard  the  right  hand  figure  of  1154, 
and  see  how  many  times  the  trial  divisor,  18,  is  contained  in  115.  The 
quotient  is  6,  which  is  probably  b,  the  tens  figure  of  the  root.  If  this  is 
correct,  (2  a  +  b),  or  the  true  divisor,  must  be  equal  to  186,  and  (2  a-}- 
b)  b  must  be  equal  to  6  times  186,  or  1116.  This  last  product,  being 
less  than  1154,  shows  that  the  work  is  correct.  We  subtract,  and  to  the 
remainder,  38  hundreds,  add  the  right  hand  period,  44  units,  which  gives 
3844  for  a  new  dividend. 

{/.)  Now,  if  we  let  a'  represent  the  part  of  the  root  already  found, 
i.  e.,  the  96  tens,  and  b'  the  units,  a'  -}-  b'  will  repretsent  the  required 
root,  and  (a'  -j-  b')2  =  a'2  _|_  (2  a'  -f  ^')  ^'  the  given  number. 

(7.)  But  we  have  already  subtracted  a'2;  the  remainder,  3844,  mast 
rontain  the  (2  a  -|-  b')  b'. 


332  RULE  FOR  SQUARE  ROOT. 

(h.)  We  now  make  2  a',  or  192,  a  trial  divisor;  and  since  2  a'V 
must  be  tens,  we  omit  the  right  hand  figure  of  3844,  and  see  how  many 
times  2  a',  or  192,  is  contained  in  384. 

(i.)  This  gives  b'  =  2,  =  the  probable  units  figure.  If  it  be  correct, 
the  true  divisor,  2  a'  +  b',must  equal  1922,  and  (2  a'  -{-  b')  b'  must  be 
2  X  1922.  This,  being  equal  to  3844,  shows  that  the  given  number  is  a 
perfect 'square,  and  962  is  its  root. 

(j.)  If  there  had  been  another  figure  in  the  root,  we  might  have 
represented  the  part  of  the  root  already  found  by  a",  or  by  some  other 
letter,  and  the  next  figure  by  b",  or  by  some  other  letter,  and  have  pro- 
ceeded as  before. 

{L)   The  numerical  work  would  be  written  thus  :  — 

925444  (  962  =  Root. 
81  =  a2 


2  a  +  b  =  186  )  1154 

1116  =  (  2  a-hb)  b 


2  a'  +  V  =  1922  )      3844 

3844  =  (  2  a'  -f-  b'  )  V 


>S10«    Huh  for  Square  Root,  with  Problems. 

As  a  similar  process  can  always  be  followed,  we  may  describe  the 
method  of  extracting  the  square  root  of  a  number  thus  :  — 

First.  Divide  the  given  number  into  periods  of  two  figures,  beginning 
with  the  units. 

Second.  Find  the  greatest  square  in  the  left  hand  period,  and  place  its 
root  as  the  highest  denomination  of  the  required  root. 

Tliird.  Subtract  the  square  thus  found  from  the  left  hand  period,  and  to 
the  remainder  bring  down  the  next  period,  calling  the  result  a  dividend. 

Fourth.   Double  the  part  of  the  root  already  found  for  a  trial  divisor. 

Fifth.  Se£  how  many  times  this  trial  divisor  is  contained  in  all  of  the 
dividend,  excepting  the  right  hand  figure,  and  xorite  the  quotient  as  the  next 
figure  of  the  root,  and  also  place  it  at  the  right  of  the  trial  dimsor^  to  form 
a  true  divisor. 

Sixth.  Multiply  this  true  divisor  by  the  root  figure  last  founds  and  sub- 
tract  the  product  from  the  dividend. 

Seventh.  Bri7)g  down  the  next  period  to  the  right  of  the  remainder,  to 
form  the  next  dividend. 

Eighth.  Double  the  part  of  the  root  already  found  for  a  trial  divisor^  and 
oroceed  xs  indicated  in  the  5<A,  6th,  1th,  and  8th  of  these  paragraphs. 


SQUARE  ROOT  OF  FRACTIONS.  333 

"N\Tiat  is  the  square  root  of  each  of  the  following  niim- 
bers :  — 


1.  7056. 

2.  9025. 

3.  3104644. 

4.  349281. 


5.  4137156. 

6.  22610025. 

7.  4260096. 

8.  38580769440964. 


317*    Square  Root  of  Fractions. 

(a.)  Since  the  square  of  a  fraction  equals  the  square  of  its  numera- 
tor divided  by  the  square  of  its  denominator,  the  square  root  of  a  frac- 
tion must  equal  the  square  root  of  its  numerator  divided  by  the  square 
root  of  its  denominator. 


Thus 


(6.)  If  the  numerator  and  denominator  of  a  fraction  are  not  perfect 
squares,  we  can  only  get  the  approximate  value  of  its  square  root. 

(c.)  In  such  cases,  if  the  denominator  is  not  a  perfect  square,  it  will 
be  well  to  multiply  both  terms  by  such  number  as  will  make  it  so.  This 
number  may  be  either  the  denominator,  or  the  product  of  the  prime  num- 
bers which  are  found  as  factors  in  the  denominator,  1,  3,  5,  or  any  odd 
number  of  times. 

^,         7          7    X  11         77         5  5  5    X  3         15 

Thus,  —  = = .      - —  = = =  -  - 

'11        11X11        121        12        22X3        22X32       36 

19_      19      _19X2X3_1U 

24  "~  23  X  3  ■"     24  X  32     ~"  f44" 

49  49  X  10  490 


.049  =  .0490,  or 


1000       1000  X  10        10000 


79 
(d.)  .  What  is  the  square  root  of  ^-r  ? 

o4 

e  7  ,.  •      79  79  79  X  3  X  7  1659 

Solution.    -—  = 


84        22  X  3  X  7        22  X  32  X  72        22  X  32  X  72 

40 


— ,  approxi- 


H  /i^=     /        ^659          _          V  1659 

®"^^'  V    84       V    22  X  3^  X  72""  2X3X7 

mately. 

This  differs  from  the  true  root  by  less  than  ^2- 

(e.)  Should  a  greater  degree  of  accuracy  be  required,  both  terms 
may  be  multiplied  by  such  a  square  number  as  will  make  the  denomina- 
tor sufficiently  large  to  secure  the  requisite  degree  of  accuracy. 

Thus,  multiplying  the  numerator  and  denominator  of  -r; » 

fJ    ^  22  X  32  X  7^ 


834  SQUARE    ROOT    OF    DECIMAL    FRACTIONS. 

41475 
by  52,  gives  22  x  3^  X  7^  X  5^'  *^®  approximate  root  of  which  it 

203  203 

=  rr-r,  which  differs  from  the  true  root  by  less  than 


2X3X7X5       210' 
210* 

What  is  the  square  root  of  each  of  the  following  frac- 
tions, — 


1. 

^*A? 

8. 

V 

5. 

s? 

2. 

MM? 

i. 

i? 

6. 

A? 

318.    Square  Itoot  of  Decimal  Fractions. 

(a.)  In  order  that  a  decimal  fraction  may  have  a  perfect  square  for 
its  denominator,  it  must  have  an  even  number  of  places  in  its  numer- 
ator. 

Thus,  the  denominators  of  .04,  .25,  .17,  .6561,  and  .384736  are  per- 
fect squares,  but  the  denominators  of  .4,  2.5,  .017,  .06561,  .0384736,  &e., 
are  not. 

{h.)  If  a  decimal  fraction,  the  root  of  which  is  required,  does  not 
have  an  even  number  of  decimal  places  in  its  numerator,  a  zero  must 
be  annexed,  to  make  the  number  even,  so  that  the  denominator  in  all 
cases  may  be  a  perfect  square. 

(0.)  Since  yxfr  =  yjl  =  i  =  .1,  jm^  =  ^^55=^ 

=  .01,  &c.,  it  follows  that  there  will  be  as  many  decimal  places  in  the  frae 
tional  part  of  the  root  as  there  are  times  two  decimal  places  in  the  frac- 
tional part  of  the  power.  Hence,  we  can  carry  out  the  root  to  as  many 
places  as  we  choose,  by  annexing  two  zeros  to  the  power  for  every 
additional  figure  which  we  wish  to  obtain  in  the  root. 

Required  ^/.4 

Solution.—  Since  ,Ja  —  ,JXo  =  ^.4000  =  y .400000,  we  may  hart 
the  following  written  work :  — 

.40  (  .6324 

.36 


1.23  )    400 

369 


1.262  )    3100 
2524 


1.2644  )    57600 
50576 

7024 


CUBE  ROOT. RELATION  OF  CUBE  TO  ROOT.   335 

The  exact  root  of  such  a  number  as  the  above  cannot  be  found,  for 
A  number  which  does,  not  end  witli  a  zero  cannot  have  zero  for  the 
right  hand  figure  of  its  square. 

What  is  the  square  root  of — 


1.  .0625  ? 

2.  5.625? 


3.  .8281?  I       5.     56.25? 

4.  1.6?  6.     .16? 


319.    Guhe  Root.  —  Relation  of  Cube  to  Root, 

(a.)  To  find  a  method  of  extracting  the  cube  root  of  a  number,  we 
must  make  some  preliminaiy  investigations  similar  to  those  of  213 
and  214. 

TABLE  OF  ROOTS  AND  CUBES. 

13  =  1  43  =  64  73  =  343 

23  =  8  .53=125  83  =  512 

33  =  27  63  =  216  93  =  729 

1 03  =  1 ,000  1 ,0003  =  1 ,000,000,000 

1 003  =  1 ,000,00c  1 0,0003  =  1 ,000,000,000,000 

(6.)    The  above  table  shows  that,  — 

First.  There  are  below  1000  but  9  entire  numbers  which  are  perfect 
cubes. 

Second.  The  entire  part  of  the  cube  root  of  any  number  below  1000 
will  be  less  than  10,  and  therefore  will  contain  but  one  figure ;  of  any 
number  between  1000  and  1,000,000  will  lie  between  10  and  100,  and 
therefore  will  contain  but  two  figures  ;  &c. 

How  many  figures  are  there  in  the  entire  part  of  the  root 
of  47986754  ? 

Answer. —  Since  47986754  lies  between  1,000,000  and  1,000,000,000, 
its  root  must  lie  between  the  roots  of  those  numbers,  i.  e.,  between  100 
and  1000,  and  hence  must  contain  3  figures  in  its  entire  part. 


330.    Division  into  Periods. 

(o.)  As  the  cube  of  10  is  1000,  of  100  is  1,000,000,  &c.,  it  follows 
that  the  cube  of  any  number  of  tens  will  be  some  number  of  thousands, 
of  any  number  of  hundreds  will  be  some  number  of  millions,  &c. ;  or, 
in  other  words,  that  the  cube  of  tens  will  give  units  of  no  denominatroo 
below  thousands,  the  cube  of  hundreds  will  give  units  of  no  denomina* 
tion  below  millions,  &c. 


336  METHXJD    OP   FORMING  A   CUBE. 

(b.)  Hence,  the  three  right  hand  figures  of  any  number  will  contain 
no  part  of  the  cube  of  the  denominations  above  units,  the  six  right  hand 
figures  will  contain  no  part  of  the  cube  of  those  above  tens,  &c. 

(c.)  Therefore,  if  we  should  begin  at  the  right  of  any  number,  and 
separate  it  into  periods  of  three  figures  each,  the  number  of  periods 
would  be  the  same  as  the  number  of  figures  in  its  cube  root.  The  cube 
of  the  highest  denomination  would  be  found  in  the  left  hand  period  the 
cube  of  the  two  highest  denominations  would  be  found  in  the  two  left 
hand  periods,  &c. 

1.  Separate  9876585925  into  periods,  and  explain  their 
uses. 

"  Answer.  9876585925.  The  left  hand  period,  9,  contains  all  of  the 
cube  of  the  thousands  of  the  root ;  the  two  left  hand  periods,  9876,  the 
cube  of  the  thousand   and  hundreds  ;  &c. 

Separate  each  of  the  following  numbers  into  periods,  and 
explain  their  uses  :  — 


2.  42783794. 

3.  584376423. 


4.  5847643759427. 

5.  6972842903612. 


331*    Method  of  forming  a   Cube, 

(a.)    To  find  a  law  of  universal  application,  we  will  use  the  letter  a 
to  represent  any  number  whatever,  and  b  to  represent  any  other  number. 
(6.)    Then  will  a  +  b  represent  the  sum,  and  (a  -j-  ^)^  or  (a  -f-  b) 
X  (a  +  b)  X  (a  +  b)i  the  cube  of  any  numbers  whatever. 

(c.)   But  (a  +  b)8  =  (a  +  b)2  X  (a  +  b)  =  (a2  +  2  ab  +  \r~)  X 
(a  +  b).    Now,  multiplying  a2  -f-  2  ab  +  b2  by  a  gives  aS  +  2  a2  b  + 
ab2,  and  multiplying  it  by  b  gives  a2  b  +  2  ab2  -j-  b^.    Adding  these 
results  together  gives  a^  -f  3  a2  b  +  3  ab2  -f-  b^. 
{d.)   This  work  may  be  written  out  thus  :  — 
a2  4-2ab4-b2=(a  +  b)a 
a  +  ^ 
a3-f  2a2b  +  ab2  =  a  X  (a2  +  2  ab  +  b-') 

a2  b  -I-  2  ab2  +  b3  =  b  X  (aS  -}-  2  ab  -f  V) 


a8  +  3a2  b  H-  3  ab2  -h  b3  ==  (a  +  b)  X  (a2  -f  2  a  b  -f-  b2)  =  (a  +  b)3 

(e.)  But  3  a2  b  4-  3  a  b2  -f  b3  =  3  a2  times  b  +  3  a  b  times  b  +  b« 
times  b  =  (3  a2  -f-  3  ab  +  b2)  times  b,  =  (3  a2  -f  3  ab  -|-  b2)  b. 

Again :  3  ab  +  b-  =  3  a  times  b  -f-  b  times  b  =  (3  a  -f-  b)  times 
b  =  (3  a  -f  b)  b. 

Hence,  a^  -f  3  a2  b  +  3  ab2  -f  bs  =  a^ -{-  [3  a2  4-  (3  a  +  b)  b}  X  b ; 


f 


TO    BXTBACT    THE    CtJli:    ROOT.  337 

L  e.,  the  cube  of  every  number  composed  of  two  parts  is  equal  to  the  cuke  of 
the  first  part ;  plus  the  product  obtained  by  multiplying  the  second  part  by 
the  sum  of  three  times  the  square  of  the  first  part,  plus  the  sum  of  three 
times  the  Jirst  part  plus  the  second  multiplied  by  the  second. 


223.     To  extract  the   Cube  EooL 

What  is  the  cube  root  of  259694072  ? 

Solution.  —  (a.)  Since  the  number  lies  between  1,000,000  and  1,000,- 
000,000,  its  root  must  lie  between  100  and  1000,  and  hence  must  contain 
three  figures.  Separating  it  into  periods  of  three  figures  each,  it  will 
take  the  form  259694072. 

(6.)  If  we  now  let  a  represent  the  hundreds  of  the  root  and  b  the 
tens,  it  is  evident  that  the  whole  of  a^  will  be  found  in  the  left  hand 
period,  and  of  (a  -f-  b)3  in  the  two  left  hand  periods. 

(c.)  The  greatest  cube  in  259  is  216,  the  root  of  which  is  6.  There- 
fore, 6  =  a,  =  the  hundreds  figure  of  the  root.  Subtracting  a^,  =  216 
millions,  from  259  millions  leaves  43  millions,  to  which  adding  the  next 
period  gives  43694  thousands.  We  may  regard  this  as  a  dividend,  and 
it  must  contain  [3  a^  -|-  (3  a  -f-  ^)  X  b]  b,  i.  e.,  the  remaining  part  of 
(a  +  b)3. 

{d.)  Now,  as  we  know  a,  we  can  find  3  a^,  and  make  use  of  it  as  a 
trial  divisor  to  find  b.  But  a  being  6  hundreds,  3  a-  must  equal  108 
ten-thousands,  and  as  b  is  tens.  Sab  must  be  hundred-thousands. 
Hence,  we  may  disregard  the  two  right  hand  figures  of  the  dividend, 
and  see  how  many  times  the  trial  divisor,  108,  is  contained  in  436. 

(e.)  The  quotient  being  4,  we  write  it  as  the  probable  tens  figure  of 
the  root,  and  have  next  to  complete  the  true  divisor,  3  a'^  -f"  {3a  -f-  b)  b. 
But  (3  a  -{-  b)  =  18  hundreds  -f  4  tens  =  184  tens,  and  (3  a  -f-  b)  b 
=  184  tens  multiplied  by  4  tens  =  736  hundreds.  Hence,  the  true 
divisor  3  a^  4-  (3  a  -f  b)  b  =  108  ten-thousands  +  736  hundreds  = 
11536  hundreds. 

(/)  Multiplying  this  by  4  gives  [3  a^  -f-  (3  a  -f-  b)  b]  b  =  4G144 
thousands,  which,  being  greater  than  43694  thousands,  shows  that  there 
are  not  as  many  as  4  tens  in  the  root,  and  that  b  is  less  than  4. 

{g.)  Assuming  b  =  3,  and  proceeding  as  before,  we  find  that  the 
true  divisor  3  a^  -|-  (3  a  +  b)  b  =  108  ten  thousands  -f"  549  hundreds 
«=  11349  hundreds ;  and  [3  a^  -|-  (3  a  -|-  ^J)  b]  b  =  34047  thousands, 
which,  being  less  than  43694  thousands,  shows  that  3  is  the  true  tens 
figure  of  the  root.  Subtracting  34047  thousands  leaves  9647  thousands, 
to  which  adding  the  next  period,  072  units,  gives  9647072  for  a  new 
dividend,  which  must  contain  the  remaining  part  of  the  power. 

(i.)  If  we  now  let  a'  represent  the  pai't  of  the  root  already  found, 
.  29 


388  TO  EXTRACT  THK  CUBE  ROOT. 

i.  e.,  the  63  tens,  and  b  the  units,  we  shall  have  259694072  =  a'3  -f* 
[3  a'-  +  (3  a'  4-  b')  b']  b',  and,  as  we  liave  already  subtracted  a'-^, 
9647072  will  contain  [3  a'2  -f  (3  a'  -f-  b')  b'j  V. 

(/.)  But  3  a'2  =  3  times  the  square  of  63  tens  =  11907  hundreds, 
which  may,  as  before,  be  made  use  of  as  a  trial  divisor  to  find  b'. 

As  3  a  -  is  hundreds  and  b'  is  units,  3  a'-  b'  must  be  hundreds;  hence, 
no  part  of  3  a'-  b'  can  be  found  to  the  right  of  hundreds,  and  wo  may 
disregard  the  two  right  hand  figures  of  the  dividend,  and  sec  how  many 
times  the  trial  divisor,  11907,  is  contained  in  9G470. 

(j.)  The  quotient  being  8,  we  write  8  as  the  probable  units  figure  of 
the  root,  and  complete  the  true  divisor,  3  a"^  +  (3  a  -}-  b)  b.  3  a  -f-  I 
=  189  tens  -f-  8  units  =  1898  units,  and  (3  a  +  b)  b  — -  1898  X  8  == 
15184.  Therefore,  3  a^  -f  (3  a  +  b)  b  =  11907  hundreds  +  15184 
units  =  1205884  =  the  true  divisor. 

(k.)  Multiplying  this  by  8  gives  9647072,  which  shows  that  8  is  the 
true  value  of  b',  and  638  is  the  root  required. 

Proof.  —  Sec  if  638'^  =  259694072. 

{/.)  Had  the  root  contained  another  figure,  we  might  have  taken  a'' 
to  represent  the  part  already  found,  and  b"  to  represent  the  next  figure, 
when  we  should  have,  (a"  +  b")=*  =  a"3  +  [3  a"2  -j-  (3  a"  +  b")  b"] 
b"  equal  tlie  number. 

{m.)  Much  of  the  labor  of  finding  the  trial  divisor,  3  a'2,  might  havo 
been  avoided.  For  as  a'  =  a  -f-  ^j  3  a'-^  must  equal  3  times  the  square 
of  a  4-  b,  or  3  times  (a2  -f  2  ab  +  b^)  =  3  a2  4.  6  ab  -{-  3  b^. 

(n.)  But  the  previous  trial  divisor,  3  a^  4-  (3  a  -|-  h)  b  =  3  a^  4- 
3  ab  4-  ^-,  f^"tl  the  number  which  stands  above  it  equals  (3  a  4"  ^)  b 
s=  3  ab  4-  b2 ;  hence  the  sum  of  these  equals  3  a2  4-  6  ab  4-  2  b3, 
which  only  lacks  b-  of  being  equal  to  3  a2  -f-  6  ab  4-  3  b^,  or  to  3  a'2. 
Hence,  by  squaring  b,  and  adding  it  to  this  result,  we  have  3  a'  =  3  a' 
4-6ab4-3b2. 

(o.)    The  work  may  be  written  thus  :  — 

a.b,V 
259694072  (638 
216000000     630  =  a' 


3  a2  ==  1080000  )    43694000 
(3  a  4-  b)  b  =  183  X  3  =      54900  \ 

[3  a2  -}-  (3  a  +  b)  b]  =  1134900  f  34047000  =  [3    a'    4- 

b2=         900  J  (3  a4-b)blb 

3  a2=:  1190700  )      9647072 
(3a'4-b')b'b'=:1898  X  8=      15184 

[3  a"  4-  (3  a'  4-  b')  V]  =  1205884         9647072  =  [3  a'2  4- 

(3  a'  4-  b')  b  1  b 


BULE  FOR  THE  CUBE  ROOT.  389 

(p.)  By  keeping  in  mind  the  denominations,  so  as  to  render  the  zeroi 
nnnecessarj,  we  should  have  the  following  form  :  — 

259694072  (  638 
216  =  a3 


Trial  divisor  =  3  a2  =      108      )  43694  =  Divid. 
(3  a  -I-  b)  b  =  183  X  3  =         549  ^ 

True  divisor,  3  a2 -f.  (3  a  +  b)  b,  =      11349  r  34047  =  [3  a"*  + 

b2  =  9  J  (3  a  -f  b)  bj  b 

Trial  divisor  =  3  a'2  =      11907  )  9647072  =  Divid. 
(  a'  +  V)  b'=          15184 

True  dir.  =  3  a'2  +  (3  a'  +  V)  b'  =      1205884  9647072  =  [3  a'2  -f. 

(3  a' +  V)  V]  b' 


SS3.    Rule  for  the   Cube  Root, 

As  a  similar  process  will  always  apply,  we  may  describe  the  method 
of  extracting  the  cube  root  of  a  number  thus  :  — 

First.  Divide  the  given  number  into  periods  of  three  figures  each,  begin- 
ning ivith  the  units. 

Second.  Find  the  greatest  cube  in  the  left  hand  period,  and  place  its  root 
as  the  first  figure  of  the  required  root. 

Thii'd.  Subtract  this  cube  from  the  left  hand  period^  and  to  the  remaindet 
bring  down  the  next  period,  calling  the  result  a  dividend. 

Fourih.  Find  three  times  the  square  of  the  part  of  the  root  already 
found,  and  make  it  a  trial  divisor. 

Fifth.  See  how  many  times  the  trial  divisor  is  contained  in  the  dividend^ 
excqjting  the  two  right  hand  figures,  and  write  the  quotient  as  the  next  figure 
of  the  root.  ' 

Sixth.  To  three  times  the  part  of  the  root  previouslg  found,  annex  the 
last  root  figure,  multiply  the  result  by  the  last  figure,  and  placing  the  product 
under  the  trial  divisor,  two  places  to  the  right,  add  it  to  the  trial  divisor. 
This  will  give  the  true  divisor. 

Seventh.  Multiply  the  true  divisor  by  the  last  root  figure,  placing  the 
product  under  the  dividend. 

Eighth.  Subtract  the  product  from  the  dividend,  ajid  to  the  remainder 
annex  the  next  period  for  a  new  dividend. 

Ninth.  Add  the  square  of  the  last  quotient  figure  to  the  last  true  divisor 
and  the  number  standing  oi'cr  it.  The  sum.  will  equal  three  times  the  square 
of  the  root  already  found,  and  will  be  the  second  trial  divisor. 

Tenth.   Now  proceed  as  directed  from  the  fifth  forward. 

What  is  the  cube  root  of  each  of  the  following  numbers  ?  — 


340 


CUBR  ROOT  OF  FRACTIONS. 


1.  830584. 

2.  262144. 

3.  676836152. 

4.  183250432. 


5.  432081216. 

6.  27054036008. 

7.  30225545875. 

8.  6804992375. 


334.     Cube  Root  of  Fractions. 

(a.)  The  cube  root  of  a  fraction  equals  the  cube  root  of  its  nuniera* 
tor,  divided  by  the  cube  root  of  its  denominator;  and  if  both  terms  of 
the  fraction  are  not  perfect  cubes,  only  its  approximate  root  can  be 
obtained. 

{b  )  If  in  such  cases  the  denominator  is  not  a  perfect  cube,  it  will  be 
well  to  multiply  both  terii.s  by  the  square  of  the  denominator,  or  by 
such  other  number  as  will  make  it  so.     Thus,  — 

3 

3 


3 
8  /3 


8/2  X  3-^_   3/18^ 
V    3  X  32       V    27 

4        V4X 


'3  X^ 
2 

1.   What  is  the  cube  root  of  ^  ? 

5  X  32  _  45 
""216* 


8 

-ye 


^^«'^'^"'      li  =  2-3^ 


23  X  33 

3 


Hence, 


2  J  6     '      6  '^  ^'  approximately. 


This  differs  from  the  true  root  by  less  than  ^. 

(c.)   If  a  greater  degree  of  accuracy  is  required,  both  terms  may  bft 
multiplied  by  some  perfect  cube  before  extracting  the  root.    Thus,  — 

il=    3/45   X  23^   s/360 
216 


216  X  23       V    1728 
which  differs  from  the  true  root  by  less  than  -rb"* 
What  is  the  cube  root  of — 


=       ,q  =  rri  approximately, 


12' 


2.  iJf? 

4.    T^a^y^r? 

6.    4? 

3.      T?f? 

5.    if? 

7.    IV 

330.    Cuhe  Root  of  Decimal  Fractions, 

(a.).  In  order  that  a  decimal  fraction  may  have  a  perfect  cube  for  iti 
denominator,  it  must  contain  3,  6,  9,  or  some  multiple  of  three  places  in 
its  numerator. 


BULE  FOR  EXTRACTING  A  ROOT  OF  AN  f  DEGREE.  341 

Thns,  the  denominators  of  .008,  .027,  .512,  .003,  and  .375067  are  each 
perfect  cubes,  while  the  denominators  of  .8,  .08,  .0027,  and  .56789  are  not. 

(c.)  If  a  decimal  fraction,  the  root  of  which  is  required,  does  not  con- 
tain 3,  6,  9,  or  some  exact  multiple  of  3  decimal  places,  zeros  mnst  bo 
annexed,  so  that  the  denominator  may  be  in  all  cases  a  perfect  cube. 


3 


s  /I 


V  ioou( 


(c?)  Since  y. 001=     / =  —  =  .1,  and  y. 000001  =     /  

^    '  ^  V   100       10  ^  V    lOOUOOO 

E=  —  =  .01,  &c.,  it  follows  that  there  will  be  as  many  decimal  places 

in  the  fractional  part  of  the  root,  as  there  are  times  three  decimal  places 
in  the  fractional  part  of  the  power.  Hence,  we  carry  out  the  root  to 
as  many  decimal  places  as  we  choose,  by  annexing  three  zeros  to  the 
power  for  each  additional  figure  we  wish  to  obtain  in  the  root. 

What  is  the  cube  root  to  four  places  of  decimals  of — 


1.  .37.  I       3.     1.76. 

2.  .6735.  4.     29.78. 


5.  .427. 

6.  .0007, 


336.    Rule  for  extracting  a  Root  of  any  degree. 

A  root  of  any  degree  may  he  found  as  follows  :  — 

First.  Divide  the  given  number  into  periods  of  as  many  figures  each  as 
there  are  units  in  the  index  of  tlie  required  root. 

Second.  Find  the  greatest  power  of  the  required  degree  in  the  left  hana 
period,  and  place  its  root  as  the  first  figure  of  tlie  required  root. 

Third.  Subtract  the  power  from  the  left  hand  figure,  and  to  the  remain- 
der bring  down  the  first  figure  of  the  next  period  for  a  dividend. 

Fourth.  Raise  the  part  of  the  root  already  found  to  a  power  one  degree 
less  than  the  given  power,  and  multiply  the  result  by  the  index  of  the  required 
root,  calling  the  result  a  trial  divisor. 

Fifth.  Divide  the  dividend  by  the  trial  divisor,  and  the  quotient  will  prob- 
ably be  the  next  figure  of  the  root.  To  ascertain  whether  it  is,  place  it  in 
the  root,  and  raise  the  number  thus  found  to  the  required  power.  If  the 
result  equals  the  first  two  periods  of  the  given  number,  or  is  less,  the  root 
figure  is  correct ;  but  if,  as  will  ofien  be  the  case,  it  is  greater,  the  root  figure 
is  too  large. 

Sixth.  Having  found  tlie  true  root  figure,  find  the  remainder,  and  form 
a  trial  divisor,  Sfc,  as  before. 

29* 


842 


MENSURATION. 


SECTION   XVII 


MENSURATION. 


237.    Polygons, 

(a.)   A  TRIANGLE  is  a  figure  having  three  sides  and  three  angles. 

(6.)   A  RIGHT-ANGLED  TRIANGLE  Is  a  triangle  having  a  right  angle. 

(c.)  For  definitions  of  the  angle,  rectangle,  and  square,  see 
pages  33  and  34. 

{d.)  Lines  are  parallel  when  they  lie  in  the  same  direction ;  as,  for 
instance,  the  lines  forming  the  sign  of  equality. 

(6.)  A  parallelogram  is  a  four-sided  figure,  having  its  opposite 
sides  parallel. 

(/)  A  TRAPEZOID  is  a  four-sided  figure  having  two  of  its  sides 
parallel. 

(g.)   A  POLYGON  is  a  figure  bounded  on  all  sides  by  straight  lines. 


Fig\. 


Fig.  2. 


Fig.  3. 


(h.)  Figures  1  and  2  represent  triangles  ;  figure  2  represents  a  right- 
angled  triangle  ;  figure  3  a  square  ;  figure  4  a  parallelogram  ;  figure  f>  a 
trapezoid  ;  and  all  represent  polygons. 

(t.)  Similar  figures  are  those  which  have  the  same  shape,  i.  e.,  which 
have  the  angles  of  the  one  equal  to  the  corresponding  angles  of  the 
other,  and  the  sides  about  the  equal  angles  proportional. 

(j.)  The  BASE  of  a  figure  is  the  side  on  which  it  is  supposed  to  stand. 
Any  side  of  a  figure  may  be  regarded  as  its  base.  The  side  opposite  the 
base  of  a  rectangle  or  parallelogram  is  often  called  its  upper  base. 

Thus,  in  figure  3,  A  D  is  the  lowwr,  and  D  C  the  upper  base. 


MENSURATION.  343 

(k.)  The  altitude  of  a  triangle  is  the  perpendicular  distance  from  the 
side  assumed  as  its  base  to  the  vertex  of  the  opposite  angle 

Thus,  in  figure  1,  when  M  N  is  taken  as  the  base,  the  distance  O  R 
is  the  altitude  5  in  figure  2  when  A  B  is  the  base,  A  C  is  the  altitude ; 
when  B  C  is  the  base,  A  D  is  its  altitude ;  and  when  A  C  is  the  base, 
A  B  is  the  altitude. 

{/.)  The  ALTITUDE  of  &  rectangle,  a  parallelogram,  or  a  trapezoid  is 
the  perpendicular  distance  between  its  parallel  bases. 

Thus,  A  B  is  the  altitude  of  figure  3,  z  q  is  the  altitude  of  figure  4, 
and  u  V  of  figure  5. 

(/«.)  A  DIAGONAL  of  a  polygon  is  a  line  drawn  from  the  vertex  of 
two  angles  lying  opposite  to  each  other. 

Thus,  B  D  and  B  C  are  diagonals  of  figure  3,  z  y  and  x  p  are  diag- 
onals of  figure  4,  F  ^,  F  J,  G  E,  &c.,  are  diagonals  of  figure  6. 

(n.)  The  area  of  a  square  or  of  a  rectangle  equals  its  length  multi- 
plied by  its  breadth.  —  See  40,  (c),  Note,  and  163,  Note. 

v(o.)    The  area  of  a  triangle  equals  half  the  product  of  its  base  by  its 
altitude. 

(p.)  The  area  of  a  parallelogram  equals  the  product  of  its  base  by 
its  perpendicular  height. 

(q.)  The  area  of  a  trapezoid  equals  half  the  product  of  its  altitude 
by  the  sum  of  its  parallel  bases. 

(r.)  The  area  of  an  irregular  polygon  can  be  found  by  dividing  it 
into  triangles. 

'"■     (s.)   The  areas  of  different  triangles,  squares,  and  parallelograms  are 
to  each  other  as  the  product  of  their  bases  by  their  altitudes. 

(t.)  The  areas  of  similar  polygons  are  to  each  other  as  the  squares 
of  their  like  dimensions. 

(w.)  The  circumference  of  a  circle  equals  very  nearly  3.1416*  times 
its  diameter. 

(v.)  The  circumferences  of  circles  are  to  each  other  as  their  diame- 
ters or  radii. 

(w.)  The  area  of  a  circle  equals  half  the  product  of  its  circumference 
by  its  radius,  or  quarter  the  product  of  its  circumference  by  its  diameter, 

(x.)  The  area  of  a  circle  also  equals  the  square  of  its  radius  multi- 
plied by  3.1416. 

(y.)  The  areas  of  circles  are  to  each  other  as  the  squares  of  their 
diameters  or  radii. 

*  More  accurately,  3,141592653589 ;  but  the  above  is  sufficiently  exact 
for  most  purposes.  Indeed,  Sy  is  sometimeB  used  where  no  great  degree 
of  accuracy  is  required. 


344  MENSURATION. 

2S8.    Problems. 

General  Direction.  —  Draw  figures  to  correspond  tc 
the  conditions  of  each  problem. 

1.  What  is  the  area  of  a  triangle  of  which  the  base  is  8  ft. 
and  the  altitude  6  ft.  ? 

2.  What  is  the  area  of  a  parallelogram  of  which  the  base 
is  9  ft.  and  the  altitude  6  ft.  ? 

3.  What  is  the  area  of  a  trapezoid  of  which  the  upper 
base  is  6  ft.,  the  lower  10  ft.,  and  the  altitude  9ft.? 

4.  What  is  the  circumference  of  a  circle  of  which  the 
diameter  is  1 6  ft.  ? 

5.  What  is  the  diameter  of  a  circle  of  which  the  circum- 
ference is  25  ft.  ? 

6.  What  is  the  area  of  a  circle  of  which  the  radius  is  4  ft.  ? 

7.  What  is  the  area  of  a  circle  of  which  the  circumference 
is  22  ft.  ? 

8.  What  is  the  radius  of  a  circle  of  which  the  area  is 
527  ft.  ? 

9.  How  many  rods  long  is  the  side  of  a  field  which  con- 
tains 40  acres  ? 

10.  How  many  rods  in  diameter  is  a  circular  field  which 
contains  8  acrcG  ? 

11.  What  must  be  the  length  of  the  side  of  a  square  field 
which  is  equal  in  area  to  a  circular  field  100  rods  in  circum- 
ference ? 

12.  A  rectangular  field  containing  100  square  rods  is  twice 
as  long  as  it  is  wide.     What  is  its  length  ? 

13.  What  must  be  the  diameter  of  a  circular  field  which 
contains  4  times  as  much  surface  as  a  similar  field  32  rods  in 
diameter  ? 

14.  The  diameter  of  one  circular  field  is  twice  that  of  an- 
other.    How  do  their  areas  compare  ? 

15.  How  many  square  feet  in  a  board  15  feet  long,  2  feet 
wide  at  one  end,  and  1  foot  wide  at  the  other,  allowing  that 
the  sides  taper  re>gularly  ? 


PROFERTIES    OF    THE    RIGHT-ANGLED    TRIANGLE.      345 

329.    Properties  of  the  Right-angled  Triangle,  with  Prolh 

lems. 

Square  of  the  Htpothenuse. 

(a.)    The  side  opposite  the  right  angle  of  a  right-angled  triangle  is 
called  the  hypothenuse. 

In  figure  2  the  side  A  C  is  the  hypothenuse. 
(6.)  The   square  of  the  hypothenuse   equals 
the    sum   of    the    squares    of    the    other    two 
sides. ' 

(c.)  The  annexed  figure  will  illustrate  the 
meaning  of  this.  Its  truth  can  be  rigidly 
demonstrated  by  geometry. 

ABC  represents  a  right-angled  triangle 
right-angled  at  B.  Let  A  B  be  4  ft,  and  B  C 
3  ft  Then  will  A  C  be  5  ft,  and  A  B^  -f  B  C^ 
==  A  C,  or  16  +  9  =  25. 

1.    What  is  the  hypothenuse  of  a  right-angled  triangle  of 
which  one  side  is  6  ft.  and  the  other  11  ft.  ? 


A  r     I  I 


Suggestion.  —  The  hypothenuse  equals  ^36  -\-  121  =  ^157  ft. 

2.  What  is  the  third  side  of  a  right-angled  triangle  of 
which  the  hypothenuse  is  12  ft.  and  the  given  side  9  fl.  ? 

Suggestion.      ^144  —  81  =  ,^63  =  Ans. 

m 

3.  What  is  the  distance  from  one  corner  of  a  floor  to  the 
opposite  corner,  if  the  floor  is  24  ft.  long  and  18  ft.  wide  ? 

4.  What  is  the  diagonal  of  a  square  30  ft.  on  a  side  ? 

0.  A  boy,  flying  his  kite,  found  that  he  had  let  out  845 
yards  of  string,  and  that  the  distance  from  where  he  stood  to 
a  point  directly  under  the  kite  was  676  yards.  How  high 
was  the  kite  ? 

6.  A  certain  window  is  20  ft.  from  the  ground.  How  long 
must  a  ladder  be  which,  having  its  foot  15  ft.  from  the  bottom 
of  the  building,  will  just  reach  the  window  ? 

7.  How  long  is  a  side  of  the  greatest  square  which  can  be 
inscribed  in  a  circle  3  feet  in  diameter  ? 


Note.  —  The  diagonals  of  a  square  bisect  each  othp^r  at  right  angles. 


346  SOLIDS. 

8.  A  tower,  60  feet  high,  stands  on  a  mound  30  feet  above 
a  horizontal  plane.  On  this  plane,  and  directly  south  of  the 
tower,  is  a  spring,  and  directly  east  from  the  spring,  and  at  a 
distance  of  1 60  feet  from  it,  on  the  same  plane,  stands  a  large 
oak  tree.  Now,  allowing  that  the  distance  in  a  direct  line 
from  the  top  of  the  tower  to  the  spring  is  150  feet,  what  is 
the  distance  from  the  top  of  the  tower  to  the  foot  of  the  oak  ? 

9.  Two  men  started  from  the  same  place,  and  travelled,  one 
n^jTth  at  the  rate  of  4  miles  per  hour,  and  the  other  east  at 
the  rate  of  3  miles  per  hour.  After  travelling  7  hours,  they 
turned,  and  travelled  directly  towards  each  other  at  the  same 
rate  as  before,  till  they  met.  How  many  miles  did  each 
travel  ? 

10.  There  is  a  rectangular  field  100  rods  long  and  80  rods 
wide,  the  sides  of  which  run  north  and  south.  A  man  started 
from  the  south-west  corner,  and  travelled  due  north  along  the 
western  boundary  of  the  field  for  60  rods,  when  he  travelled 
across  the  field  in  a  straight  line  to  the  north-east  corner. 
How  much  farther  did  he  travel  than  he  would  if  he  had 
gone  in  a  straight  line  all  the  way  ? 

330.    Solids. 

(a.)  A  SPHERE  is  a  solid  bounded  by  a  curved  surface,  every  pffft  of 
which  is  equally  distant  from  a  point  within,  called  the  centre. 

(b.)  A  line  drawn  from  the  centre  to  the  surface  is  called  a  radius, 
and  a  line  drawn  from  any  point  in  the  surface  through  the  centre  to  the 
opposite  point  is  called  a  diameter. 

(c.)  A  PRISM  is  a  solid  having  its  several  faces  parallelograms,  and 
its  bases  two  equal  and  parallel  polygons. 

(d.)   A  CUBE  (see  41)  6.)  is  a  kind  of  prism. 

(e.)  A  CYLINDER  is  such  a  solid  as  would  be  formed  by  revolving  a 
rectangle  about  one  of  its  sides.  It  has  also  been  defined  to  be  "  a 
^  round  body  with  circular  ends." 

(/)  A  PYRAMID  is  a  solid  body  bounded  laterally  by  triangles,  of 
which  the  vertices  meet  at  a  common  point,  and  the  bases  terminate  in 
the  sides  of  a  polygon,  which  forms  the  base  of  the  pyramid. 

(</.)  A  CONE  is  a  solid  which  has  a  circular  base,  and  tapers  regularly 
to  a  point  called  the  vortex. 


SOLIDS. 


347 


(k.)  A  FRUSTUM  of  a  cone  or  pyramid  is  a  part  cut  off  by  a  plane 
parallel  to  the  plane  of  its  base. 

(«.)  Similar  solids  have  the  same  shape,  i.  e.,  the  angles  of  one  of 
them  equal  the  corresponding  angles  of  the  other,  and  the  sides  about 
the  equal  angles  are  proportional. 

All  spheres  are  similar.  Two  cones,  or  two  cylinders,  are  similar 
when  J;heir  altitudes  are  to  each  other  as  the  radii  or  diameters  of  their 
bases. 

Fig.  1.  Fig.  2.  Fig.  3. 


f\\ 

P\ 

i 
1 

j. 

f^ig.4. 


Fig.  5. 


(/)  Figure  1  represents  a  sphere;  figure  2  a  prism;  figure  3  a  cyl- 
inder ;  figure  4  a  pyramid  ;  figure  5  a  cone ;  figure  6  a  frustum  of  a 
cone. 

{k.)  The  SURFACE  of  a  sphere  equals  the  square  of  its  diameter 
multiplied  by  3.1416.* 

(/.)  The  surfaces  of  sphere|  are  to  each  other  as  the  squares  of  their 
radii  or  diameters. 

(m.)  The  solidity,  or  solid  contents,  of  a  sphere  ecjuals  the 
product  of  the  surface  multiplied  by  ^  of  the  radius,  or  by  ^  of  the 
diameter,  or  it  equals  ^  of  the  cube  of  the  diameter  multiplied  by 
3.1416.* 

(n.)  The  solidities  of  spheres  are  to  each  other  as  the  cubes  of  their 
radii  or  diameters. 


*  See  foot  note,  page  343. 


348  SOLIDS. 

(o.)  The  solidities  of  similar  solids  are  to  each  other  as  the  cubes  of 
their  like  dimensions. 

(p.)  The  solidity  of  a  prism  equals  the  area  of  its  base  multiplied  by 
4ts  altitude. 

(q.)  The  solidity  of  a  cylinder  is  equal  to  the  area  of  its  base  multi- 
plied by  its  altitude. 

(r )  The  convex  surface  of  a  cylinder  is  equal  to  the  circumference 
of  its  base  multiplied  by  its  altitude. 

(s.)  The  solidity  of  a  cone  or  of  a  pyramid  equals  the  area  of  its 
base  multiplied  by  -j  of  its  altitude. 

(t.)  The  solidity  of  a  frustum  of  a  cone,  or  of  a  pyramid,  equals  ^ 
of  the  product  of  its  altitude  multiplied  by  the  sum  of  its  upper  base, 
plus  its  lower  base,  plus  the  mean  proportional  between  the  two  bases. 

Note.  —  The  mean  proportional  of  two  numbers  is  the  square  root 
of  their  product.     Thus,  the  mean  proportional  of  4  and  9  =  ^/i  X  9 

331.    Prohlems. 

1.  What  is  the  solidity  of  a  sphere  the  diameter  of  wliich 
is  3  feet  ? 

2.  What  is  the  surface  of  a  sphere  the  radius  of  which  is 
1  foot? 

3.  What  is  the  diameter  of  a  sphere  of  which  the  solidity 
is  10  feet? 

4.  What  is  the  circumference  of  a  sphere  the  solidity  of 
which  is  12  feet? 

5.  What  is  the  diameter  of  a  sphere  of  wliich  the  surface 
is  6  feet  ? 

6.  What  is  the  solidity  of  a  prism  of  which  the  altitude  is 
9  feet,  and  the  base  contains  10  square  feet? 

7.  What  is  the  solidity  of  a  cylinder  of  which  the  altitude 
is  6  feet  and  the  radius  of  the  base  2  feet  ? 

8.  What  is  the  convex  surface  of  a  cylinder  of  which  the 
diameter  of  the  base  is  5  feet  and  the  altitude  4  feet  ? 

9.  What  is  the  solidity  of  a  cone  of  which  the  altitude  is  9 
feet  and  the  circumference  of  the  base  10  feet? 

10.  What  must  be  the  diameter  of  a  sphere  whicl;  contains 
8  times  as  many  cubic  feet  as  one  3  feel  ir.  diameter  ? 


PROGRESSIONS.  349 

SECTION   XVIII. 

PROGRESSIONS. 

232,    Arithmetical  Progressiori. 

(«.)  A  SERIES  OF  NUMBERS  IN  ARITHMETICAL  PROGRES- 
SION, or  an  arithmetical  series,  is  a  series  of  numbers 
each  of  which  differs  from  the  preceding  by  the  same  number. 

(h.)  Such  a  series  would  be  obtained  by  continually  adding 
the  same  number  to,  or  subtracting  it  from,  any  given  number. 

Thus  we  should  have  — 

By  adding  2's  to  1, I,  3,  5,  7,  9,  11,  &c. 

By  adding  7's  to  3, 3,  10,  17,  24,  31,  38,  45,  &c. 

By  subtracting  4's  from  29,  .  .  .  25,  21,  17,  13,  9,  &c. 
By  subtracting  3's  from  56,  .  .  .  53,  50,  47,  44,  41,  &c. 

(e.)  If  the  series  is  formed  by  addition,  it  is  called  an 
INCREASING  SERIES  ;  if  by  Subtraction,  it  is  called  a  de- 
creasing SERIES. 

(c?.)  The  numbers  composing  a  series  are  called  the  terms 
of  the  series. 

(e.)  The  difference  between  the  consecutive  terms  of  any  series  ia 
called  the  common  difference,  and  is  always  the  number  by  the 
addition  or  subtraction  of  which  the  series  is  formed. 

(/)  Since  the  terms  of  a  series  are  formed  by  continual  additions  or 
subtractions  of  the  same  number,  it  follows  that  the  second  term  of  any 
series  equals  the  first,  plus  or  minus  the  common  difference  ;  that  the 
th'.rd  equals  the  first,  plus  or  minus  twice  the  common  difference  ;  that 
the  fourth  term  equals  the  first,  plus  or  minus  three  times  the  common 
difference ;  &c. 

{g.)  Hence,  any  term  of  an  arithmetical  series  is  equal  to  the  first  term, 
plus  or  minus  the  common  difference  taken  one  less  times  than  there  are  terms 
in  the  series  ending  with  the  required  term. 

(h.)  Moreover,  if  the  first  term  of  an  increasing  arithmetical  series  he 
subtracted  from  the  last,  or  if  the  last  term  of  a  decreasing  series  be  sub- 
tracted from  the  first,  the  remainder  will  be  the  product  of  the  common  dif- 
ference multiplied  by  one  less  than  tJie  number  of  terms. 

80 


350  PROGRESSIONS. 

233.    ProUems. 

1.  What  is  the  10th  term  of  the  increasing  series  of 
which  the  first  term  is  3  and  the  common  difference  8  ? 

2.  "What  is  the  25th  term  of  the  decreasing  series  of  which 
the  first  term  is  85  and  the  common  difference  2  ? 

3.  What  is  the  common  difference  of  the  series  of  which 
the  1st  term  is  7  and  the  13th  term  43  ? 

4.  How  many  terms  are  there  in  the  series  of  which  the 
1st  term  is  8,  the  last  term  85,  and  the  common  difference  7  ? 

5.  What  is  the  common  difference  of  the  series  of  which 
596  is  the  1st  term  and  491  the  22d  ? 

6.  How  many  terms  are  there  in  the  series  of  which  12  is 
the  first  term,  4  the  last,  and  \  the  common  difference  ? 

334.    To  find  the  Sum  of  a  Series. 

(a.)  If  we  should  invert  any  series,  we  should  have  a  new  one,  which 
would  differ  from  the  former  only  in  the  order  of  its  terms,  the  one  be- 
ing an  increasing  while  the  other  is  a  decreasing  series.  The  first  term 
of  one  series  would  equal  the  last  of  the  other,  and  each  term  of  one 
peries  would  be  as  much  greater  than  its  preceding  terra  as  each  of  th^ 
other  is  less  than  its  preceding  term.  Hence,  if  we  should  write  the  two 
series  under  each  other,  and  add  together  the  corresponding  terms  in 
the  order  in  which  they  stand,  the  successive  sums  would  equal  each 
other,  and  each  would  equal  the  sum  of  the  first  and  last  terms  of  the 
original  series. 

(b.)  Moreover,  there  would  be  as  many  such  sums  as  there  are  terms 
in  the  series.  Hence,  the  sum  of  the  two  series,  or  (since  they  are 
equal)  twice  the  sum  of  either  of  them,  is  equal  to  the  product  obtained 
by  multiplying  the  first  and  last  terms  by  the  number  of  terms. 

Thus,  by  inverting  the  series  3,  7,  9,  &c.,  to  35,  we  have,  — 
3       7     11     15     19     23     27     31     35  =  given  series. 
35    31     27    23     19     15    11       7      3  =  same  series  inverted. 


38    38     38     38     38    38    38    38     38  =  the  sums  of  the  succes- 
sive terms. 

(er.)  Adding  these  last  results  together  would  give  the  sum  of  the 
two  series,  or  twice  the  sum  of  either,  which  would  manifestly  be  equal 
to  9  times  38,  or  9  times  the  sum  of  the  first  and  last  terms.  Dividing 
this  by  2  would  give  the  sum  of  one  of  the  series. 


GEOMETRICAL    PROGRESSION.  351 

{d.)  Hence,  the  sum  of  a  series  in  arithmetical  progression  equals  half 
the  product  obtained  by  multiplying  the  sum  of  the  first  and  last  terms  by  ths 
number  of  terms. 

335.   Problems. 

1.  What  is  the  sum  of  the  series  of  which  9  is  the  1st 
term  and  94  the  20th  ? 

Answer.     (£  +  9i)j<^  =  io3  X  10  =  1030. 

2.  What  is  the  sum  of  the  series  of  which  427  is  the  1st 
term  and  187  the  81st  ? 

3.  What  is  the  sum  of  the  series  of  which  4  is  the  1st 
term  and  9  is  the  6th  ?     What  is  the  common  difference  ? 

4.  What  is  the  sum  of  the  series  of  which  the  1st  term  is 
7,  the  common  difference  9,  and  the  number  of  terms  15  ? 

5.  How  many  terms  are  there  in  a  series  of  "which  the 
sum  is  648,  the  1st  term  3,  and  the  last  term  78  ?  What  is 
the  common  difference  ? 

6.  What  is  the  1st  term  and  common  difference  of  a 
series  of  which  the  last  term  is  164,  the  number  of  terms  12, 
and  the  sum  2100  ? 

7.  Form  the  series  of  which  the  sum  is  153,  the  1st  term 
1,  and  the  last  term  17  ? 

336.     Geometrical  Progression. 

(a.)  A  series  of  numbers  in  geometrical  progression,  or  a 
geometrical  series,  is  a  series  of  numbers  each  of  which  bears 
the  same  ratio  to  the  one  which  follows  it. 

(b.)  Such  a  series  would  be  obtained  by  continually  multi- 
plying or  dividing  by  the  same  number. 

Thus,  beginninj^  with  2  and  multiplying  by  3,  we  should  have  2,  6, 
18,  54,  162,  486,  &c. 

By  beginning  with  3072  and  multiplying  by  ^,  we  have  3072,  1536, 
768,384,  192,  96,  48,  &c. 

(c.)    The   numbers   comprising  the   series   are   called  the 

TERMS    OF    THE    SERIES. 


S52  GEOMETRICAL    PROGRESSION. 

(d.)  The  ratio  of  each  term  to  that  which  follows  it  13 
called  the  common  ratio,  and  is  always  the  number  by 
which  we  multiplied  to  produce  the  series. 

(e.)  If  it  be  an  increasing  series,  the  common  ratio  will  equal  a  whole 
number  or  an  improper  fraction ;  but  if  it  be  a  decreasing  series,  the 
common  ratio  will  equal  a  proper  fraction. 

(/.)  From  the  method  of  forming  such  series,  it  is  obvious  that  the 
second  term  must  equal  the  first  multiplied  by  the  common  ratio  ;  that 
the  third  term  must  equal  the  first  multiplied  by  the  second  power  of 
the  common  ratio  ;  «Ssc. 

(g.)  Hence,  any  term  of  a  geometrical  series  must  equal  the  product  of  the 
first  term  multiplied  by  the  common  ratio  raised  to  a  power  one  degree  lest 
than  the  number  of  the  term. 

{h.)  Moreover,  if  the  last  term  of  a  geometrical  series  be  divided  by  the 
first,  the  quotient  will  be  the  common  difference  raised  to  a  power  one  degree 
less  than  the  number  of  the  term. 

237.    ProUems, 

1.  What  is  the  7th  term  of  the  series  of  which  125  is  the 
1st  term,  and  2  the  common  ratio  ? 

2.  What  is  the  5th  term  of  the  series  of  which  1  is  the 
1st  term  and  ^  the  common  ratio  ? 

3.  What  is  the  9th  term  of  the  series  of  which  4096  is  the 
1st  term  and  ^  the  common  ratio  ? 

4.  What  is  the  common  ratio  of  the  series  of  which  1  is 
the  1st  and  81  the  5th  term  ? 

5.  Construct  the  series  of  which  1  is  the  1st  and  6561  is 
the  9  th  term. 

6.  Construct  a  series  of  8  terms,  having  1  for  the  1st 
term  and  §  for  the  common  ratio. 

238.    To  find  the  Sum  of  a   Geometrical  Series, 

(a.)  If  each  term  of  a  geometrical  series  should  be  multiplied  by  the 
common  ratio,  a  new  scries  would  be  formed,  of  which  the  first  term 
would  equal  the  second  term  of  the  former  series,  the  second  term  would 
equal  the  third  of  the  former,  &c. ;  the  last  term  but  one  of  the  new 
series  would  equal  the  last  of  the  given  series.  Hence,  the  first  term  of 
the  original  series  would  have  no   corresponding  term  in  the  derived 


GEOMETRICAL    PROGRESSION.  353 

series,  and  the  Ifist  term  of  the  derived  series  would  have  no  correspond* 
ing  term  in  the  original  series. 

Thus,  by  multiplying  each  term  of  the  series  2,  6,  18,  &c.,  to  1458  by 
the  common  ratio,  we  have  — 

2     6     18     54     162    486     1458  =  given  series. 

6     18     54     162     486     1458     4374  =  derived  scries  =  3  times 
given  series. 

Again.    Multiplying  each  term  of  the  series  32,  8,  2,  -,  &c.,  to  — —  by 

2  128 

tlie  common  ratio  -,  we  have  — 
4 

32     8     2    -    -    —-    — —  =  given  series. 
2     8     32     128       ^ 

o«lll        1         1  ,.,.  I,. 

8    2    -    -    —  —  —  =  derived  series  =  -  of  given 

2     8     32  128  512  4  ^ 

series. 

(6.)  Now,  as  the  given  series  equals  once  itself,  and  the  derived  series 
equals  the  common  ratio  times  the  given  series,  it  follows  that  the  dif- 
ference between  the  given  and  derived  series  will  equal  the  product  of 
the  given  series  multiplied  by  the  difference  between  1  and  the  common 
ratio. 

(c.)  But,  as  has  been  shown,  the  difference  between  the  series  equals 
the  difference  between  the  first  term  of  the  given  series  and  the  last 
term  of  the  derived  series. 

(d.)  Hence,  to  find  the  sum  of  a  geometrical  series,  we  may  multiply  t^ 
last  term  by  the  common  ratio,  and  divide  the  difference  between  the  product 
and  the  first  term  by  the  difference  between  \  and  the  common  ratio. 

239.   ProUems. 

1.  What  is  the  sum  of  the  series  of  wliich  2  is  the  1st 
term,  1458  the  last,  and  3  the  common  ratio  ? 

Solution.  3  times  1458  =  4374,  from  which  subtracting  2  leaves 
4372.  Dividing  this  by  3  —  1,  or  2,  gives  2186  for  the  sum  of  the 
series. 

2.  What  is  the  sum  of  the  series  of  which  32  is  the  1st 
term,  ^^^  the  last,  and  ^  the  common  ratio  ? 

Solution.  T^^  X  i  =  ttjWj  which  taken  from  32  leaves  31x^14- 
This  equals  1  —  ^,  or  f ,  times  the  required  sum.  Hence,  the  sum  of 
the  series  equals  Sly^ff  -5-  f  =  "^^ris- 

3.  What  is  the  sum  of  the  series  of  which  ^  is  the  1st 
term,  486  the  last,  and  3  the  common  ratio  ? 

30* 


854  CIRCULATING    DECIMALS. 

4.  What  is  the  sum  of  a  series  of  11  terms  of  which  16  w 
the  1st  term  and  ^  the  common  ratio  ? 

5.  What  is  the  sum  of  a  series  of  5  terms  of  which  5  is 
the  1st  term  and  3125  the  last  term  ? 

340.    Infinite  Decreasing  Series. 

(a.)  As  in  a  decreasing  series  each  terra  is  smaller  than  the  preced- 
ing, it  follows  that  if  the  series  be  carried  far  enough,  the  terms  will 
become  so  small  that  they  may  be  disregarded  without  affecting  sensibly 
the  sum  of  the  §eries. 

(6.)  An  infinite  decreasing  series  will  always  be  of  this  character, 
and  hence  its  sum  will  equal  the  quotient  obtained  by  dividing  the  first 
term  by  the  difference  between  1  and  the  common  ratio. 

1.  What  is  the  sum  of  the  infinite  decreasing  series  of 
which  4  is  the  1st  term  and  ^  the  common  ratio  ? 

Answer.    4-f-{l  —  ^)  =  4  -r-  ^  =  4-5. 

What  is  the  sum  of  the  infinite  decreasing  series  — 

2.  Of  which  1  is  the  1st  term  and  ^  the  common  ratio  ? 

3.  Of  which  3  is  the  1st  term  and  f  the  common  ratio  ? 

4.  Of  which  2  is  the  1st  term  and  |  the  common  ratio  ? 

5.  Of  which  .37  is  the  1st  term  and  .01  the  common 
ratio  ? 

6.  Of  which  .597  is  the  1st  term  and  .001  the  common 
ratio  ? 

7.  Of  Fhich  .2794  is  the  1st  term  and  .0001  the  common 
ratio  ? 


SECTION    XIX. 
S41.    CIRCULATING  DECIMALS. 

(a.)  A  circulating  or  repeating  decimal  is  one  which  will 
never  terminate,  but  in  which  the  same  figure,  or  succession 
of  figures,  will  always  follow  each  other  in  the  same  order. 


CIRCULATING    DECIMALS.  856 

Examples.     .9999,  &c.;  .323232,  &c. ;  .5174351743,  &C.; 

.0200602006,  &c. 

(h.)  Circulating  decimals  are  equivalent  to  vulgar  frac- 
tions, the  exact  decimal  value  of  which  cannot  be  found.  — 
See  143,  («.) 

(c.)  That  such  vulgar  fractions  must  give  rise  to  repeating  decimals 
may  be  shown  thus  :  — 

Since  the  remainder  after  any  division  must  be  less  than  the  divisor, 
we  shall  at  some  stage  of  the  division,  explained  in  143,  find  a  remain- 
der equal  to  a  former  remainder,  and  from  this  point  the  quotients  and 
remainders  will  succeed  each  other  in  the  same  order  as  before. 

(d.)  A  repeating  decimal  is  indicated  by  placing  a  dot  ovei 
the  repeating  figure,  or  over  the  first  and  last  figures  of  the 
repeating  period. 

Thus,  .7  =  .7777,  &c. ;  .19  =  .191919,  &c. , 

.453142  =  .4531423142,  &c. 

(e.)  The  repeating  part  of  a  decimal  will  begin  as  soon  as  all  the 
factors  of  10  have  been  cancelled  from  the  denominator  of  the  vulgar 
fraction  which  produces  it,  the  vulgar  fraction  being  in  all  cases  reduced 
to  its  lowest  terms.  —  See  143,  (h.) 

1.  With  which  place  will  the  repetend  of  the  decimal 
value  of  2^  begin  ? 

Answer.  —  Since  12  contains  the  third  power  of  2,  which  is  a  factor 
of  10,  the  repetend  will  commence  after  3  places  have  been  obtained, 
i.  e.,  with  the  fourth  place. 

2.  "With  which  place  will  the  repetend  of  ^  commence  ? 

Answer.  —  Since  24  contains  no  factor  of  10,  the  repetend  will  com- 
mence with  the  first  place. 

With  which  place  will  the  repetend  of  each  of  the  follow- 
ing^ fractions  commence  ? 


3.    f. 
4-    H' 


7-    iJ. 


8.     tV 


5.  if. 

6.  iff. 
(f.)   An  expression  which  contains  only  the  figures  of  the  repetend 

is  called  a  single   repetend  ;   one  which  contains  other  figures  is 
called  a  mixed  repetend. 

Thus,  .427  is  a  simple  repetend,  and  .53427  is  a  mixed  repetend. 


35B  CIRCULATING    DECIMALS. 

(g.)  Two  repetends  are  similar  when  they  begin  at  the  same  decimal 
place,  and  conterminous  when  they  end  at  the  same  decimal  place. 

TJius,  .523  and  .9,  or  .2473,  and  .417  are  similar;  .427  And  .436,  or 
.4279,  and  0372  are  conterminous. 

(h.)  The  repeating  period  may  be  considered  as  beginning  at  anjr 
figure,  provided  that  it  is  made  to  include  the  entire  combination  which 
is  repeated. 

-     Thus,  .417  =  .4174  =  .41741  =  .417417 ;  for  each  developed  would 
give  .417417417417,  &c. 

{i.)  A  repeating  decimal  is  really  an  infinite  decreasing  series  in 
geometrical  progression,  of  which  the  first  term  is  the  first  repeating 
period,  and  the  common  ratio  is  the  decimal  fraction,  having  1  for  a 
numerator  and  the  power  of  10,  whose  exponent  contains  as  many  units 
as  there  are  places  in  the  repeating  period,  for  its  denominator.  ( See 
last  three  problems  in  239.) 

Thus,  .48  =  a  series  of  which  .48  is  the  1st  term  and  .01  the  com 
mon  ratio.     Hence,  .48  =  .48  ^  (1  —  .01 )  =  .48  -i-  .99  =  ||. 

Again.  .479  =  a  series  of  which  .479  is  the  1st  term  and  .001  the 
common  ratio.     Hence,  .479  ==  .479  -r-  (1  —  .001)  =  .479  -^  .999  == 

HI 

So  ^298  =  .3298  -4-  (1  —  .0001)  =  .3298  -h  .9999  =  f Iff. 

(/.)   We  should  reach. the  same  result  by  observing  that  — 

^  =  .111,  &c.  =  .1  ;  ^V  == -010101,  &c.  =  .01  ; 

^^^  =  .001001,  &c.  =  .OOi  ;  ^^V^  =  .00010001,  &c.  =  .0001  ;  &c 

Now,  .5  ==  5  X  .i  =1 ;  .8  ==  8  X  .1=1; 

.27  =  27  X  .6i  =  f  i- ;  .49  =  49  X  .51  =  || ; 

.147  =  147  X  .OOl  =  Ml-;  -045  =  45  X  -OOl  =  -^Vg-,  &c. ; 

.4657  =  .4f  H  ;  .328923  =  .32|||f . 

(Jc.)  Hence,  it  follows  that  every  repeating  decimal  is  equivalent  to  a 
tulgar  fraction,  of  which  the  numerator  is  expressed  by  the  repeating 
period,  and  the  denominator  by  as  many  9's  as  there  are  figures  in  the 
repeating  period. 

What  is  the  value  of — 

1.  .87?       I     3.  .3006?     I     5.  .5252  1 

2.  .4864?      I     4.  .2794?     I     6.  .237982? 

{l.y  In  multiplying  a  repeating  decimal  by  any  multiple  or  power  of 
ten,  care  must  be  taken  to  fill  the  places  left  vacant  by  the  change  of 
the  point  by  the  figures  of  the  repeating  period,  and  to  observe  what 
figures  would  have  to  be  added  to  any  period  on  account  of  the  multi- 
plicati(m  of  the  preceding  period. 


CIRCULATING    DECIM  VLS.  857 


Thus,  .9  X  1000  =  99.9       .345  X  10  =  3.453 

2.743  X  30  =  2.743743,  &C.,  X  30  =  82.312 


What  is  the  product  of — 

1.     .4  X  100  ? 

4. 

..307  X  10000? 

2.     27.1436  X  4000? 

5. 

28.43  X  1000  1 

3.     .84635  X  200000  1 

6. 

.1374  X  8000  1 

(m.)  Repeating  decimals,  like  other  fractions,  are  of  the  same  de« 
nomination  only  when  they  are  fractional  parts  of  the  same  unit,  and 
have  a  common  denominator. 

(«.)  This  is  the  case  only  when  they  are  similar  and  conterminous. 
To  make  circulating  decimals  similar  and  conterminous,  then,  is  to 
reduce  them  to  the  same  denomination,  which  may  be  done  by  carrying 
them  out  so  far  that  repeating  periods  of  the  same  number  of  figures 
beginning  and  ending  at  the  same  place  may  be  formed  in  each. 

1.  Make  .3587  and  .423  similar  and  conterminous. 

Answer.     .35878787  and  42342342,  i.  e.,  .35|J||-|^  and  .42f|-f fff. 

Make  the  repetends  in  each  of  the  following  cases  similar  and  con- 
terminous :  — 


2.  .24  and  .375. 

3.  .467  and  .52. 

4.  .1784  and  .328. 


5.  .5182  and  4.16. 

6.  .6153  and  .4i37. 

7.  94287  and  51281. 


(o.)  In  adding  repetends,  or  in  multiplying  them,  care  must  be  taken 
to  observe  what  figures  should  be  added  to  the  written  periods,  on  ac- 
count of  the  addition  or  multiplication  of  the  next  lower  periods. 

(p.)    Thus,  in  adding  .247  +  -639  +  -587',  we  may  observe  that  since 
*  the  sum  of  the  left  hand  figures  is  13,  there  must  be  1  to  bring  from  the 
next  period  below.    Adding  this  with  the  period  gives  1.474. 

1.   What  is  the  product  of  .239  X  8  ? 

Answer.  —  Observing  that  8  times  39  gives  3  of  the  next  higher  de- 
nomination, we  have  8  X  3.239  =  1.9i5. 


What  is  the  value  of — 

2.  .47  +  .325  +  .i  7 -f- .327  ? 

3.  .6279  +  .3284568  ? 

4.  ..579  X  7  ? 

5.  69.432746  X  37  ? 


6.  .5183 +  4.61  4-3.85  + .217' 

7.  52.17697  +  1.38463? 

8.  29.476  X  12  ? 

9.  43.006  X  8  1 


Circulating  decimals  are  multiplied  and  divided  by  each  other 
vulgar  fractions  are. 


858  MISCELLANEOUS    EXAMPLES. 


What  is  the  value  of — 

1.  .679  X  .4  ? 

2.  .279  X  .5286  ? 

3.  .68  -4-  .i7  1 

4.  5763  -f-  .02 1 


42.76  X  .437  1 
23.84  X  27.96  ? 
2.974  -i-  .3007  1 
1.728  -^  .0i44  ? 


SECTION  XX. 


24-3.    Miscellaneous  Examples. 

1.  How  many  pounds  of  iron  are  equal  in  weight  to  100  pounds  of 
gold? 

2.  4^  times  a  certain  number  added  to  1  equals  y  of  the  quotient  ob- 
tained by  dividing  6f  by  if.     What  is  the  number  1 

3.  How  many  seconds  are  there  from  9f  o'clock,  A.  M.,  of  June  4, 
1855,  to  2^  o'clock,  P.  M.,  of  Aug.  7,  1855  ? 

4.  John  can  do  a  piece  of  work  in  5  days,  and  James  can  do  it  in  6 
days.     How  many  days  will  it  take  both  to  do  it  ? 

5.  How  many  hogsheads,  of  63  gallons  each,  will  a  cylindrical  cis- 
teiTi,  8  feet  in  diameter  and  10  feet  high,  hold  1 

6.  What  is  the  least  common  multiple  and  the  greatest  common 
divisor  of  74333,  313131,  171717,  146853,  and  .53095? 

7.  What  is  the  value  of  77  +  ^  ? 

8.  Mr.  Taylor  has  invested  \  of  his  money  in  railroad  stock,  f  of  it 
in  bank  stock,  ^  of  it  in  real  estate,  and  the  rest  in  trade.  Moreover, 
what  he  has  invested  in  railroad  stock  is  $20,000  more  than  he  has  in- 
vested in  trade.  How  much  money  has  he,  and  how  much  has  he 
invested  in  each  way  ? 

9.  Moses  E.  Fuller  bought  18  shares  of  bank  stock  at  an  advance  of  S 
per-oent  on  their  par  value  of  $100  per  share.  Six  months  afterwards, 
and  at  the  end  of  every  subsequent  6  months,  he  received  a  dividend  of 
44-  per  cent.  At  the  end  of  2  yr.  3  mo.  he  sold  the  stock  at  a  premium 
of  12  per  cent.  Money  being  worth  6  per  cent  per  year,  compound  in- 
terest, how  much  did  he  gain  by  the  speculation  ? 

10.  Mr.  Goodwin  and  Mr.  Brown  commence  trade  with  equal  sums  of 
money.    Mr.  Goodwin  gained  $2000,  and  Mr.  Brown  lost  10  percent 


WISCELLANEOUS    EXAMPLES. 


359 


of  his  stock,  when  it  was  found  that  Mr.  Goodwin  had  just  twice  as 
much  as  Mr.  Brown.     What  was  the  original  stock  of  each  ? 

11.  A  young  man  having  contracted  a  debt  equal  to  f  of  his  income, 
found  that,  by  saving  ^5-  of  his  income  annually,  he  could  in  5  years 
pay  up  his  debt  and  have  $50  left.     What  was  his  income  ? 

12.  The  hour  and  minute  hands  of  a  watch  are  together  at  12  o'clock 
When  will  they  next  be  together  ? 

13.  At  what  time  between  12  and  1  will  the  hour  and  minute  hands 
of  a  watch  point  in  opposite  directions  ? 

14.  What  was  due  on  the  following  account,  July  1,  1855,  interest 
being  6  per  cent  per  year  ? 


Dr. 


Wm.  Barnes,  in  account  with  James  Shedd. 


Cr. 


1855. 

1  1855. 

Jan.  17. 

To  Sund.,  6  mo. 

$673 

42 

Feb.  1. 

By  Sund.,  4  mo. 

$237 

80 

Jan.  28. 

To  Sund.,  4  mo. 

542 

31 

Mar.  5. 

By  Sund.,  3  mo. 

492 

5C 

Feb.  13. 

To  Sund.,  6  mo. 

237 

23 

Mar.  21. 

Bv  Sund.,  6  mo. 

873 

27 

Apr.  22. 

To  Sund.,  3  mo. 

720  60 

lApr.  25. 

By  Sund.,  3  mo. 

594 

82 

May  10. 

To  Sund.,  2  mo. 

54  20 

jMay  27. 

By  Sund.,  4  mo. 

376 

15 

June  23. 

To  Sund.,  3  mo. 

133  60 

June  2. 

By  Sund.,  2  mo. 

142 

60 

! 

;June  20. 

By  Sund.,  6  mo. 

225 

00 

1 5.  What  is  the  value  of  a  pile  of  wood  40  ft.  long,  4  ft.  wide,  and  5 
ft.  high,  at  $5.30  per  cord  ? 

16.  A  owes  B  $144,  due  in  6  mo.  20  da.,  and  B  owes  A  $324,  due  in 
1  yr.  4  mo.  and  20  da.  If  A  should  pay  half  of  his  debt  now,  and  the 
other  half  when,  by  the  conditions,  the  whole  debt  was  due,  when  ought 
B  to  pay  the  whole  of  his  1 

17.  A  owes  B  $600  dollars,  due  in  9  mo.,  and  B  owes  A  $900,  due  in 
15  mo.  If  A  does  not  pay  his  debt  till  B's  would  othei'wise  have  be- 
come due,  when  ought  B,  injustice,  to  pay  his  debt  to  A? 

18.  A  man  travelled  100  miles  in  two  days.  ^  of  the  distance  he 
travelled  the  first  day,  added  to  ^  the  distance  he  travelled  the  second 
day,  equals  ^  the  distance  he  travelled  the  first  day.  How  far  did  he 
travel  each  day  1 

19.  A  set  out  from  Providence  to  go  to  Boston,  a  distance  of  42 
miles,  and  B  at  the  same  time  left  Boston  for  Providence.  At  the  end 
of  six  hours  they  met,  when  it  appeared  that  A  had  travelled  l^-  miles 
per  hour  more  than  B.     How  far  had  each  travelled  ? 

20.  A  dishonest  silversmith  bought  a  bar  of  gold  at  $192  per  lb.,  and 
sold  it  for  $16  per  ounce,  weighing  it  in  both  cases  by  avoirdupois 
weight.  How  much  did  he  gain  by  the  fraud,  allowing  that  the  true 
%veight  of  the  bar  was  5  pounds,  and  that  gold  was  worth  $16  pel 
ounce  ? 


860  MISCELLANEOUS    EXAMPLES. 

21.  A  certain  room  is  16  ft.  long,  15  ft.  wide,  and  12  ft.  high.  What 
is  the  distance  from  the  right  hand  upper  corner  to  the  left  hand  lower 
corner  ? 

22.  A  vessel  of  war  left  port  with  provisions  enough  to  last  600  men 
18  months.  At  the  end  of  three  months  she  captured  and  sunk  an 
enemy's  vessel,  and  took  on  board  her  crew  of  150  men.  Two  months 
after  she  captured  another  vessel,  on  board  which  she  placed  the  prison- 
ers taken  from  the  former  prize,  and  100  men  besides.  Four  riionths 
after  she  captured  another  vessel,  on  board  which  she  placed  50  men. 
When  she  returned  to  port  again  she  had  provisions  enough  to  last  the 
crew  which  was  left  on  board  of  her  1  month.  How  long  was  she  ab- 
sent from  port  ? 

23.  A  prize  of  $31,000  is  to  be  distributed  among  three  officers  and 
eight  sailors,  so  that  each  officer  shall  have  2^  times  as  much  as  a  sailor. 
What  will  be  the  share  of  each  ? 

24.  Take  any  number  whatever,  multiply  it  by  3,  add  7,  subtract  the 
original  number,  multiply  by  2,  add  6,  divide  by  4,  add  27,  subtract  the 
original  number,  and  the  result  is  32.     Why  is  this  ? 

25.  A  merchant  sold  50  bushels  of  wheat  for  Mr.  Kandall,  and  60 
Dushels  for  Mr.  Palmer,  receiving  $150  for  the  lot.  Now,  allowing  that 
Mr.  Randall's  was  worth  20  per  cent  more  per  bushel  than  Mr.  Palmer's, 
how  ought  the  money  to  be  divided  ? 

26.  I  bought  25,000  feet  of  boards  at  $2.25  per  thousand,  and  sold  ^ 
of  them  for  what  §  of  them  cost.  What  per  cent  did  I  gain  on  the  part 
sold  1 

27.  I  bought  63  kegs  of  nails,  each  keg  containing  100  pounds,  at  4^ 
cents  per  pound,  and  sold  §  of  them  for  what  ^  of  them  cost.  What 
per  cent  did  I  lose  on  the  part  sold  ? 

28.  I  sold  ^  of  a  lot  of  land  for  the  cost  of  f  of  the  lot,  and  the 
remainder  for  j  of  what  I  sold  the  first  part  for.  What  per  cent  of  its 
cost  did  I  gain  on  the  entire  lot  ? 

29.  A  certain  sum  is  to  be  divided  among  three  persons  in  such  a 
way  that  the  first  has  2  dollars  as  often  as  the  second  has  3  and  the 
third  6,  It  turns  out  that  the  third  has  12  dollars  more  than  both  the 
others.     What  was  the  sum  divided,  and  the  share  of  each  person  ? 

~  30.  A  hare  starts  30  yards  before  a  greyhound,  but  is  not  seen  by  liim 
till  she  has  been  up  20  seconds.  If  the  hare  runs  at  the  rate  of  8 
miles  per  hour,  and  the  hound  at  the  rate  of  10  miles  per  hour,  how 
long  will  the  chase  continue,  and  how  far  must  each  run  from  his  place 
of  starting? 

31.  How  many  per  cent  in  advance  of  the  cost  must  a  merchant  ask 
for  goods,  that,  after  allowing  for  a  loss  of  6  per  cent  of  his  sales  by  bad 
debts,  an  average  credit  of  6  months,  and  expenses  equal  to  8  per  cent 


MISCELLANEOUS    EXAMPLES.  361 

of  the  cost  of  the  goods,  he  may  make  a  gain  of  10  per  cent  of  the 
original  cost,  money  being  worth  6  per  cent  per  year  ? 

32.  At  a  certain  time  between  12  and  1  the  minute  hand  lacked  as 
much  of  being  at  the  1  mark  as  the  hour  hand  was  beyond  the  12  mark. 
What  time  was  it  ? 

33.  A  gentleman  bought  a  horse,  chaise,  and  harness  for  $450.  The 
horse  cost  3^  times  as  much  as  the  harness,  and  ih£  chaise  cost  $50 
more  than  the  horse.    What  was  the  cost  of  each  ? 

34.  George  Rice  has  80  yards  of  broadcloth,  which  he  will  sell  for 
cash  at  $4  per  yard,  but  for  which  he  will  ask  $4.50  per  yard  in  exchange 
for  other  commodities.  Edward  Tyler  has  silk  for  which  he  charges 
$1.25  pel  yard,  cash.  How  much  ought  he  to  charge  per  yard  for  it,  in 
exchange  for  Rice's  broadcloth  1  If,  however,  he  should  pay  $100  cash, 
and  the  balance  in  silk,  how  many  yards  of  silk  ought  he  to  give  ? 

35.  My  agent  in  New  Orleans  has  sold  for  me  a  lot  of  goods  for 
$3375,  on  a  credit  of  6  months,  and  got  the  note  discounted  at  a  bank. 
If  he  charges  4  per  cent  for  services  in  selling,  and  3  per  cent  for  guar- 
antying the  note  at  the  bank,  how  much  ought  he  to  remit  to  me  1 

36.  July  1,  1851,  I  got  ray  note  for  $1000,  payable  in  3  months,  dis- 
counted at  a  bank,  and  immediately  invested  the  money  received  on  it 
in  land.  Oct.  7,  1851,  I  sold  the  land  at  an  advance  of  12  per  cent, 
receiving  ^  of  the  sales  in  cash,  and  a  note  for  the  other  half,  payable 
July  1, 1852,  without  grace,  and  to  be  on  interest  at  7  per  cent  after  Jan. 
1,  1852.  I  lent  the  cash  at  6  per  cent  interest.  When  my  note  at  the 
bank  became  due.  I  renewed  it  for  the  same  time  as  before,  and  at  the 
proper  time  renewed  it  again ;  and  when  this  last  note  became  due,  I 
renewed  it  for  such  time  that  the  new  note  would  become  due  July  1, 
1852.  Allowing  that  I  paid  6  per  cent  interest  on  the  money  borrowed 
at  the  bank,  and  that  I  made  a  complete  settlement  July  1,  1852,  what 
was  the  amount  of  my  gains  1 

37.  April  16,  1850,  I  bought  of  Mr.  Curry  498  cords  of  wood  at  $3.50 
per  cord,  giving  in  payment  my  note  payable  on  demand  with  interest. 
Oct.  5,  1850,  I  sold  232  cords  of  it  at  $3.75  per  cord,  cash,  and  imme- 
diately lent  the  money  received  for  it  on  interest.  Oct.  17,  1850,  I  sold 
the  remainder  at  $4.07  per  cord,  to  be  paid  Jan.  1,  1851,  and  to  be  on 
interest  after  Nov.  1,  1850.  Jan.  1,  1851,  I  collected  the  money  due  me, 
and  paid  that  due  to  Mr.  Curry.  How  much  did  I  gain  or  lose  by  the 
transactions  ? 

38.  How  much  will  it  cost  to  plaster  the  walls  and  ceiling  of  a  room 
16  ft.  3'  long,  14  ft.  2'  wide,  and  12  ft.  high,  allowing  for  2  doors,  each  7 
ft.  high  and  3  ft.  6'  wide,  four  windows  5  ft.  high  and  2  ft.  8'  wide,  a  fire- 
place and  mantel  4  ft.  3'  high  and  5  ft.  wide,  and  a  mopboard  6  inches 
wide,  provided  that  it  costs  1 1  cents  per  yard. 

31 


362  MISCELLANEOUS    EXAMPLES. 

3S.  An  importer  sold  cloth  to  a  wholesale  dealer,  and  gamed  10  per, 
cent  of  what  it  cost  him.  The  wholesale  dealer  sold  it  to  a  retail  dealer 
at  an  advance  of  10  per  cent  on  what  it  cost  him.  The  retail  dealer 
sold  it  at  an  advance  of  20  per  cent  on  what  it  cost  him.  Now,  allow- 
ing that  the  retail  dealer  received  $726  for  the  cloth,  how  much  did  it 
cost  the  importer  1 

40.  What  must  be  the  diameter  of  a  globe  which  contains  27  times 
as  many  cubic  inches  as  a  globe  2  inches  in  diameter  ? 

41.  A  man  spent  ^  and  -g-  of  his  money,  and  then  earned  $36,  when 
he  had  $88  more  than  ^  of  what  he  had  at  first.  How  much  money 
had  he  at  first  1 

42.  George  says  that  if  he  should  eani  as  much  more  money  as  he 
now  has,  5-  as  much  more,  and  372-  dollars,  he  should  have  $555.  How 
much  money  has  he  ? 

43.  Four  men  bought  a  grindstone  4  feet  in  diameter,  paying  equal 
sums,  and  they  agreed  that  the  first  should  grind  off  his  share,  then  the 
second,  and  so  on  to  the  last.  What  was  the  thickness  of  the  portion 
ground  off  by  each  ? 

44.  June  1,  1852,  I  bought  for  cash  500  casks  of  oil,  each  cask  con- 
taining 42  gallons,  at  $1.10  per  gallon.  Oct.  1,  1852,  I  sold  it  on  3 
months'  credit,  at  a  price  per  gallon  equal  to  125  per  cent  of  its  cost  per 
gallon,  deducting  5  per  cent  of  the  whole  quantity  of  oil  for  leakage. 
I  immediately  got  the  note  received  for  the  oil  discounted  at  a  bank. 
Allowing  that  I  paid  10  cents  per  cask  for  truckage,  and  $25  for  storage 
and  other  expenses,  and  that  money  was  worth  6  per  cent  per  year,  did 
I  gain  or  lose,  and  how  many  dollars  ? 

45.  A,  B,  and  C  trade  in  company,  and  gain  $100,  of  which  A  has 
$12.50,  B  has  $25,  and  C  has  $62.50,  C  put  in  $21  more  of  the  origi- 
nal stock  than  A  and  B  together.     What  was  the  original  stock  ? 

46.  Reuben  Aldrich  and  George  Guild  bought  cloth  together.  Aid- 
rich  paying  $6  more  than  ^  of  its  cost.  They  sold  the  cloth  at  such 
rate  that  Aldrich's  share  was  $39,  and  Guild's  share  was  $36.  What 
did  the  cloth  cost  1 

47.  Charles,  John,  and  James  were  talking  of  their  money.  Charles 
has  50  dollars.  James  says  that  if  Charles  should  give  him  his  money, 
he  should  have  twice  as  much  as  John ;  and  John  says  that  if  Charles 
should  give  him  his  money,  he  should  hare  three  times  as  much  as 
James,     How  much  has  each  ? 

48.  A  speculator  borrowed  $2000,  agreeing  to  pay  interest  at  the 
rate  of  9  per  cent  per  year,  and  invested  the  money  in  land  at  $80 
per  acre.  3  mo.  afterwards  he  sold  ^  the  land  for  $900,  and  the  rest 
at  $100  per  acre,  and  expended  the  proceeds  for  flour.  2  mo,  15  da. 
afterwards  he  sold  ^  of  the  flour  for  $600,  and  the  remainder  for  what 


MISCELLANEOUS    EXAMPLES.  363 

he  paid  for  the  whole,  and  immediately  paid  the  amount  of  the  bor- 
rowed money.    How  much  was  his  gain  1 

49.  If  ^  of  12  were  6,  what,  in  the  same  ratio,  would  ^  of  50  be  ? 

50.  I  own  a  square  lot  of  land  measuring  10  rods  on  a  side.  How 
deep  a  ditch  6  feet  wide  must  I  dig  around  it  within  its  limits  to  raise  its 
surface  1  foot? 

51.  A  merchant  sold  a  lot  of  flour  at  $8.40  per  barrel,  and  thereby 
gained  20  per  cent.  He  afterwards  sold  another  lot  of  the  same  flour 
for  $203,  and  thereby  gained  16  per  cent.  How  many  barrels  were  there 
in  the  last  lot  ? 

52.  What  must  be  the  diameter  of  a  sphere  to  contain  as  many  cubic 
inches  as  a  cone  1  foot  high  and  having  a  base  of  1  foot  in  diameter  1 

53.  Mr.  Hicks  invested  a  certain  sum  in  flour,  and  Mr.  Gardiner  in- 
vested twice  as  much.  It  turned  out  that  Mr.  Hicks  lost  10  per  cent, 
and  Mr.  Gardiner  gained  10  per  cent,  and  that  the  difference  between 
what  Mr.  Hicks  received  for  Ms  lot  and  what  Mr.  Gardiner  received  for 
his  was  $260.    How  much  did  each  invest  ? 

54.  Jan.  1,  18.52,  I  borrowed  $954,  agreeing  to  pay  interest  at  the 
rate  of  5  per  cent,  and  immediately  expended  it  for  cloth  at  $3  per  yard. 
Four  days  afterwards  I  sold  the  cloth  at  $3.50  per  yard,  to  be  paid  June 
17,  1852.  On  receipt  of  the  money,  I  immediately  expended  it  for 
cloth  at  $1  per  yard.  July  1,  1852,  I  sold  the  cloth  at  $1.12^  per  yard, 
payable  Sept.  22,  1852.  As  soon  as  this  debt  was  paid,  I  put  the  money 
on  interest  at  6  per  cent.  Jan.  1,  1853,  I  collected  the  amount  due  me, 
and  paid  that  which  I  owed.  How  much  had  I  gained  by  the  transac- 
tions ? 

55.  Wishing  to  find  the  distance  between  two  trees,  which  cannot 
be  directly  measured  on  account  of  a  swamp,  I  measure  due  east  80 
yards  from  the  foot  of  one  of  them ;  then  turning  south,  I  measure 
100  yards,  when  I  find  that  I  am  just  40  yards  to  the  east  of  the  other 
tree.     How  far  apart  are  the  trees  1 

56.  The  eaves  of  a  house  are  at  the  same  height,  and  30  feet  apart. 
The  ridge  pole  is  12  feet  higher  than  the  eaves,  and  just  midway  be- 
tween them.  The  house  is  40  feet  long.  How  many  shingles  will  it 
take  to  cover  the  roof,  if  each  shingle  covers  a  space  6  inches  long 
and  4  inches  broad  f 

57.  Multiply  any  number  by  4,  add  6,  divide  by  2,  add  7,  divide  by  2, 
subtract  5,  add  twice  the  original  number,  divide  by  the  original  num- 
ber, subtract  1,  multiply  by  the  original  number,  add  3,  multiply  by  3, 
add  11,  divide  by  2,  and  the  result  will  always  be  10  more  than  3  times 
the  original  number.     Why  is  this  ? 

58.  A  man  spent  ^  of  his  money  and  ^  a  dollar  more ;  then  ^  of  what 
he  had  left,  and  i"  of  a  dollar  more  j  then  ^  of  what  he  had  left,  and  ^ 


364  MISCELLANEOUS    EXAMPLES. 

of  a  dollar  more,  when  he  had  just  7  dollars.    How  much  money  had 
he  at  first  1 

59.  A  rectangular  box  is  3  times  as  long  as  it  is  wide,  and  twice  as 
wide  as  it  is  high,  and  contains  96  cubic  feet.  How  many  square  feet 
are  there  in  its  surface  ? 

60.  Obtained  at  a  bank,  on  my  note  payable  in  3  months,  money 
enough  to  buy  20  acres  of  land  at  $100  per  acre.  One  month  after- 
wards I  sold  the  land,  receiving  in  payment  a  note  on  demand,  with 
interest  at  6  per  cent  per  year.  I  collected  the  amount  of  this  note  the 
day  my  note  became  due  at  the  bank,  and  found  that  it  took  3^  of  it  to 
pay  the  latter.    For  how  much  per  acre  did  I  sell  the  land  ? 

61.  Obtained  at  a  bank,  on  my  note  payable  in  5  months,  money 
enough  to  buy  20  acres  of  land  at  $100  per  acre,  and  at  the  same  time 
hired  of  a  friend  money  enough  to  buy  another  lot  of  the  same  size  and 
price  as  the  first,  giving  in  payment  my  note  payable  on  demand,  with 
interest  at  6  per  cent  per  year.  When  the  note  at  the  bank  became  due, 
I  sold  both  lots  for  cash  at  the  same  price  per  acre,  and  found  that  the 
money  received  for  16  acres  of  it  was  sufficient  to  pay  my  note  at  the 
bank.  How  much  did  I  gain  by  the  transactions  ?  How  much  more  on 
the  second  lot  than  on  the  first  ? 

62.  Bernard  Farwell  and  Francis  Dana  traded  in  company.  Farwell's 
stock  was  $860,  and  Dana's  $420.  On  dividing  their  profits,  they  found 
that  Farwell's  share  was  $4  more  than  twice  Dana's.  How  many  dol- 
lars did  each  gain  ? 

63.  A  commission  merchant  received  on  consignment  100  bags  of 
corn  from  A,  150  bags  from  B,  and  75  bags  from  C,  and  putting  thera 
all  into  1  lot  sold  them  for  $400.  Now,  allowing  that  A's  lot  is  10  per 
cent  better  than  B's,  and  15  per  cent  better  than  C's,  what  is  the  just 
share  of  each  1 

64.  Is  the  reasoning  process  contained  in  the  following  solution  true 
or  false  1    If  false,  in  what  does  its  fallacy  consist  1 

Question.  —  What  is  the  effect  of  adding  3  to  both  numerator  and 
dehominator  of  a  fraction  ? 

Solution.  —  Since  adding  3  to  both  numerator  and  denominator  of  a 
fraction  gives  for  a  result  a  fraction  which  expresses  3  more  parts  than 
the  former,  but  of  such  kind  that  it  will  take  3  more  of  them  to  equal  a 
unit,  the  addition  has  both  increased  and  diminished  the  fraction,  and 
has  therefore  not  altered  its  value. 

65.  Show  the  fallacy,  if  any,  in  the  following  solution :  — 

If  the  numerator  of  a  fraction  is  4  greater  than  its  denominator,  what 
will  be  the  effect  of  adding  the  same  number  to  both  numerator  and  de* 
nominator  1 

Solution.  —  Adding  the  same  number  to  both  numerator  and  denomi 


MISCELLANEOUS    EXAMPLES.  865 

nator  of  a  fraction  will  not  affect  the  difference  between  them.  Hence 
the  resulting  fraction  in  the  case  supposed  must,  like  the  original  one, 
express  4  more  fractional  parts  than  it  takes  to  equal  a  unit.  Therefore. 
the  resulting  fraction  equals  the  original  one,  and  the  supposed  addition 
has  not  affected  the  value  of  the  fraction. 

66.  Two  men,  A  and  B,  hired  a  horse  and  carriage  for  $7,  to  go  from 
Providence  to  Boston  and  back,  the  distance  between  the  cities  being  42 
miles.  At  Attleboro',  12  miles  from  Providence,  they  took  in  C,  agree- 
ing to  take  him  to  Boston  and  back  to  Attleboro'  for  his  proportionate 
sliare  of  the  expense.  At  Walpole,  24  miles  from  Providence,  they  took 
in  D,  agreeing  to  take  him  to  Boston  and  back  to  Walpole  for  his  pro- 
portionate share  of  the  expense.    What  ought  each  person  to  pay  1 

Note.  —  Arithmeticians  do  not  agree  as  to  the  correct  solution  of 
such  examples  as  the  above ;  some  contending  that  each  person  should 
pay  in  exact  proportion  to  the  number  of  miles  he  rode,  and  others  that 
A  and  B  should  each  pay  ^  the  expense  of  the  ride  from  Providence  to 
Attleboro',  -g-  the  expense  of  the  ride  from  Attleboro'  to  Walpole,  and  ^ 
the  expense  of  the  ride  from  Walpole  to  Boston ;  that  C  should  pay  ^ 
the  expense  of  the  ride  from  Attleboro'  to  Walpole,  and  :|-  the  expense 
of  the  ride  from  Walpole  to  Boston ;  and  that  D  should  pay  only  i 
of  the  expense  of  the  ride  from  Walpole  to  Boston.  Which  is  the  just 
principle  ? 

67.  I  sold  f  of  a  lot  of  land  for  20  per  cent  more  than  it  cost,  and  the 
remainder  for  20  per  cent  less  than  it  cost.  What  per  cent  did  I  gain  on 
the  whole  ? 

68.  I  sold  ^  of  a  cask  of  wine  for  $36,  which  was  25  per  cent  more 
than  the  part  sold  cost.  I  then  sold  the  remainder  at  an  advance  of  20 
per  cent  on  its  cost.    What  per  cent  of  the  cost  of  the  cask  did  I  gain  ? 

69.  I  sent  to  my  agent  in  Boston  a  lot  of  flour,  which  he  sold  for 
$6075,  charging  a  commission  of  2  per  cent  on  the  sales.  He  invested 
the  remainder,  after  deducting  his  commission  of  1^  per  cent  on  the 
purchase,  in  cloths,  which  he  shipped  to  my  agent  in  Savannah.  The 
latter  sold  them  at  an  advance  of  25  per  cent  on  the  cost,  charging  a 
commission  of  5  per  cent  on  the  sales,  and  invested  the  balance,  after 
deducting  a  commission  of  2  per  cent  on  the  purchase,  in  cotton.  The 
cotton  was  shipped  to  my  agent  in  Boston,  who  sold  it  at  an  advance  of 
20  per  cent  on  its  cost,  charging  a  commission  of  if  per  cent.  Allow- 
ing that  the  expenses  of  freight,  insurance,  &c.,  were  $1000,  what  was 
my  gain,  supposing  that  the  flour  cost  me  $6075  ? 

70.  A  traveller  had  to  pass  three  toll  gates.  At  the  first  gate  he  paid 
5  cents  less  than  half  the  money  he  had  ;  at  the  second  he  paid  2  cents 
less  than  half  of  what  he  had  left ;  and  at  the  third  he  paid  1  cent  more 

31* 


366  MISCELLANEOUS    EXAMPLES. 

than  half  of  A^hat  he  then  had,  after  which  he  had  only  4  cents  left 
How  much  money  did  he  have  at  first  ? 

71.  Lyman  Richards  and  John  Dexter  traded  in  company,  Eicharda 
paying  in  $9  less  than  f  of  the  whole  stock.  They  gained  $200,  of 
which  Richard's  share  was  $117.    What  was  the  original  stock  of  each  "^ 

72.  A  farmer  had  his  sheep  in  three  pastures.  In  the  first  pasture  there 
were  twice  as  many  as  in  the  second,  and  iir  the  second  twice  as  many 
as  in  the  third.  40  jumped  out  of  the  first  pasture  into  the  second, 
and  32  jumped  from  the  second  into  the  third,  when  the  number  of  sheep 
in  each  pasture  was  equal.    How  many  were  originally  in  each  pasture  ? 

73.  A  water  tub  holds  73  gallons.  The  pipe  which  conveys  water  to 
it  admits  7  gallons  in  5  minutes,  and  the  tap  discharges  20  gallons  in  17 
minutes.  Now,  suppose  that  both  being  carelessly  left  open,  the  water 
is  turned  on  at  4  o'clock,  and  a  servant  discovers  it  at  6,  and  puts  in  the 
tap,  at  what  time  will  the  tub  be  full  ?  * 

74.  A  gentleman  has  two  horses,  and  a  carriage  worth  £1 00.  Now, 
if  the  first  horse  be  harnessed  in  the  carriage,  he  and  the  carriage 
together  will  be  worth  three  times  as  much  as  the  second  horse  ;  but  if 
the  second  horse  be  harnessed,  he  and  the  carriage  will  together  be 
worth  7  times  as  much  as  the  first.    What  is  the  value  of  each  horse  ?  * 

75.  There  is  a  fish  whose  head  is  10  feet  long ;  his  tail  is  as  long  as 
his  head  and  half  of  his  body,  and  his  body  is  as  long  as  his  head  and 
tail  together.    What  is  the  whole  length  of  the  fish  1  * 

76.  "  If  12  oxen  will  eat  3^  acres  of  grass  in  4  weeks,  with  all  that 
grows  during  that  time,  and  21  oxen  eat  10  acres  in  9  weeks,  with  all 
that  grows  during  that  time,  how  many  oxen  would  eat  24  acres,  with 
the  growth,  in  18  weeks,  the  grass  all  the  while  growing  uniformly  ?  " 

77.  Jan.  1,  1851,  my  agent  in  Buffalo  bought  for  me  1000  barrels  of 
flour  at  $4  per  barrel,  for  which  he  charged  a  commission  of  1  per  cent. 
On  the  3d  of  January,  I  sent  him  cash  to  pay  for  the  flour  and  his  com- 
mission. It  cost  me  $1  per  barrel  to  have  the  flour  transported  to 
Boston,  and  I  incurred  other  expenses  upon  it  to  the  amount  of  $20. 
Feb.  1,  1851,  I  sold  the  flour  to  J.  Smith  &  Co.,  at  an  advance  of  25  per 
cent  on  its  entire  cost,  receiving  in  payment  half  cash,  and  their  note 
payable  in  6  months  for  the  remainder.  I  had  their  note  discounted  at 
a  bank ;  but  before  it  became  duo  they  failed,  so  that  when  it  became 
due,  I,  as  indorser,  was  obliged  to  pay  it.  Jan.  1,  1852, 1  settled  with  J. 
Smith  &  Co.,  receiving  50  cents  on  each  dollar  they  owed  me.  Allow- 
ing that  money  was  worth  6  per  cent  per  year,  and  that  I  paid  the 
freight  and  other  expenses  of  the  flour  on  the  15th  of  January,  what 
was  the  amount  of  my  loss  ? 


•  From  Pike's  Arithmetic,  edition  of  1788. 


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